Mixing
Why does L'Hopital's rule fail in this case?
$begingroup$
$$lim_{x to infty} frac{x}{x+sin(x)}$$
This is of the indeterminate form of type $frac{infty}{infty}$, so we can apply l'Hopital's rule:
$$lim_{xtoinfty}frac{x}{x+sin(x)}=lim_{xtoinfty}frac{(x)'}{(x+sin(x))'}=lim_{xtoinfty}frac{1}{1+cos(x)}$$
This limit doesn't exist, but the initial limit clearly approaches $1$. Where am I wrong?
calculus limits limits-without-lhopital
edited Mar 24 '16 at 18:06
Barry Cipra
60.5k655129
asked Mar 23 '16 at 21:08
Andrew FountAndrew Fount
7421612
$endgroup$
|
show 10 more comments
$begingroup$
$$lim_{x to infty} frac{x}{x+sin(x)}$$
This is of the indeterminate form of type $frac{infty}{infty}$, so we can apply l'Hopital's rule:
$$lim_{xtoinfty}frac{x}{x+sin(x)}=lim_{xtoinfty}frac{(x)'}{(x+sin(x))'}=lim_{xtoinfty}frac{1}{1+cos(x)}$$
This limit doesn't exist, but the initial limit clearly approaches $1$. Where am I wrong?
calculus limits limits-without-lhopital
edited Mar 24 '16 at 18:06
Barry Cipra
60.5k655129
asked Mar 23 '16 at 21:08
Andrew FountAndrew Fount
7421612
$endgroup$
75
$begingroup$
If the limit of $f'/g'$ exists, then it is also the limit of $f/g$. Not the other way around.
$endgroup$
– user251257
Mar 23 '16 at 21:11
8
$begingroup$
One often forgets there are hypotheses to check before applying L'Hospital. One of these is that the ratio of the derivatives must exist (or still be indeterminate).
$endgroup$
– Bernard
Mar 23 '16 at 21:13
3
$begingroup$
A condition on the use of L'Hôpital in this context is that the derivative of the denominator must be non-zero on $(N, infty)$ for some $N$.
$endgroup$
– Brian Tung
Mar 23 '16 at 21:15
2
$begingroup$
This post might give you something to think about.
$endgroup$
– Hirshy
Mar 23 '16 at 21:23
6
$begingroup$
An excellent example for a Calculus Course.
$endgroup$
– dwarandae
Mar 24 '16 at 4:46
|
show 10 more comments
$begingroup$
$$lim_{x to infty} frac{x}{x+sin(x)}$$
This is of the indeterminate form of type $frac{infty}{infty}$, so we can apply l'Hopital's rule:
$$lim_{xtoinfty}frac{x}{x+sin(x)}=lim_{xtoinfty}frac{(x)'}{(x+sin(x))'}=lim_{xtoinfty}frac{1}{1+cos(x)}$$
This limit doesn't exist, but the initial limit clearly approaches $1$. Where am I wrong?
calculus limits limits-without-lhopital
edited Mar 24 '16 at 18:06
Barry Cipra
60.5k655129
asked Mar 23 '16 at 21:08
Andrew FountAndrew Fount
7421612
$endgroup$
$$lim_{x to infty} frac{x}{x+sin(x)}$$
This is of the indeterminate form of type $frac{infty}{infty}$, so we can apply l'Hopital's rule:
$$lim_{xtoinfty}frac{x}{x+sin(x)}=lim_{xtoinfty}frac{(x)'}{(x+sin(x))'}=lim_{xtoinfty}frac{1}{1+cos(x)}$$
This limit doesn't exist, but the initial limit clearly approaches $1$. Where am I wrong?
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited Mar 24 '16 at 18:06
Barry Cipra
60.5k655129
asked Mar 23 '16 at 21:08
Andrew FountAndrew Fount
7421612
edited Mar 24 '16 at 18:06
Barry Cipra
60.5k655129
asked Mar 23 '16 at 21:08
Andrew FountAndrew Fount
7421612
edited Mar 24 '16 at 18:06
Barry Cipra
60.5k655129
edited Mar 24 '16 at 18:06
Barry Cipra
60.5k655129
edited Mar 24 '16 at 18:06
Barry Cipra
60.5k655129
60.5k655129
asked Mar 23 '16 at 21:08
Andrew FountAndrew Fount
7421612
asked Mar 23 '16 at 21:08
Andrew FountAndrew Fount
7421612
asked Mar 23 '16 at 21:08
Andrew FountAndrew Fount
7421612
7421612
75
$begingroup$
If the limit of $f'/g'$ exists, then it is also the limit of $f/g$. Not the other way around.
$endgroup$
– user251257
Mar 23 '16 at 21:11
8
$begingroup$
One often forgets there are hypotheses to check before applying L'Hospital. One of these is that the ratio of the derivatives must exist (or still be indeterminate).
$endgroup$
– Bernard
Mar 23 '16 at 21:13
3
$begingroup$
A condition on the use of L'Hôpital in this context is that the derivative of the denominator must be non-zero on $(N, infty)$ for some $N$.
$endgroup$
– Brian Tung
Mar 23 '16 at 21:15
2
$begingroup$
This post might give you something to think about.
$endgroup$
– Hirshy
Mar 23 '16 at 21:23
6
$begingroup$
An excellent example for a Calculus Course.
$endgroup$
– dwarandae
Mar 24 '16 at 4:46
|
show 10 more comments
75
$begingroup$
If the limit of $f'/g'$ exists, then it is also the limit of $f/g$. Not the other way around.
$endgroup$
– user251257
Mar 23 '16 at 21:11
8
$begingroup$
One often forgets there are hypotheses to check before applying L'Hospital. One of these is that the ratio of the derivatives must exist (or still be indeterminate).
$endgroup$
– Bernard
Mar 23 '16 at 21:13
3
$begingroup$
A condition on the use of L'Hôpital in this context is that the derivative of the denominator must be non-zero on $(N, infty)$ for some $N$.
$endgroup$
– Brian Tung
Mar 23 '16 at 21:15
2
$begingroup$
This post might give you something to think about.
$endgroup$
– Hirshy
Mar 23 '16 at 21:23
6
$begingroup$
An excellent example for a Calculus Course.
$endgroup$
– dwarandae
Mar 24 '16 at 4:46
75
75
$begingroup$
If the limit of $f'/g'$ exists, then it is also the limit of $f/g$. Not the other way around.
$endgroup$
– user251257
Mar 23 '16 at 21:11
$begingroup$
If the limit of $f'/g'$ exists, then it is also the limit of $f/g$. Not the other way around.
$endgroup$
– user251257
Mar 23 '16 at 21:11
8
8
$begingroup$
One often forgets there are hypotheses to check before applying L'Hospital. One of these is that the ratio of the derivatives must exist (or still be indeterminate).
$endgroup$
– Bernard
Mar 23 '16 at 21:13
$begingroup$
One often forgets there are hypotheses to check before applying L'Hospital. One of these is that the ratio of the derivatives must exist (or still be indeterminate).
$endgroup$
– Bernard
Mar 23 '16 at 21:13
3
3
$begingroup$
A condition on the use of L'Hôpital in this context is that the derivative of the denominator must be non-zero on $(N, infty)$ for some $N$.
$endgroup$
– Brian Tung
Mar 23 '16 at 21:15
$begingroup$
A condition on the use of L'Hôpital in this context is that the derivative of the denominator must be non-zero on $(N, infty)$ for some $N$.
$endgroup$
– Brian Tung
Mar 23 '16 at 21:15
2
2
$begingroup$
This post might give you something to think about.
$endgroup$
– Hirshy
Mar 23 '16 at 21:23
$begingroup$
This post might give you something to think about.
$endgroup$
– Hirshy
Mar 23 '16 at 21:23
6
6
$begingroup$
An excellent example for a Calculus Course.
$endgroup$
– dwarandae
Mar 24 '16 at 4:46
$begingroup$
An excellent example for a Calculus Course.
$endgroup$
– dwarandae
Mar 24 '16 at 4:46
|
show 10 more comments
5 Answers
5
active
oldest
votes
$begingroup$
Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule. What the rules says is IF the limit of $f'$ over $g'$ exists then the limit of $f$ over $g$ also exists and the two limits are the same. It doesn't say anything if the limit of $f'$ over $g'$ doesn't exist.
answered Mar 23 '16 at 21:17
Barry CipraBarry Cipra
60.5k655129
$endgroup$
67
$begingroup$
Not just a common error, a VERY common error. And I am from now on going to use the OP's example in my next calc batch!
$endgroup$
– imranfat
Mar 23 '16 at 21:18
6
$begingroup$
What's even funnier is examples where you apply l'Hopital twice and get back where you started :-)
$endgroup$
– gnasher729
Mar 25 '16 at 21:11
add a comment |
$begingroup$
L'Hopital's rule only tells you that if the modified limit exists and has value $L$, then the original limit also exists and has value $L$. It doesn't tell you that the converse holds.
So, the fact that the modified limit doesn't exist gives you no information about the original limit. So, you need a different method.
Consider something more direct: can you compute
$$
lim_{xtoinfty}frac{x}{x+sin x}=lim_{xtoinfty}frac{1}{1+frac{sin x}{x}}?
$$
answered Mar 23 '16 at 21:12
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Or 1 - sin x / (x + sin x).
$endgroup$
– gnasher729
Mar 24 '16 at 22:16
2
$begingroup$
or squeeze theorem using -1 <= sin x <= 1.
$endgroup$
– djechlin
Mar 25 '16 at 18:57
add a comment |
$begingroup$
Because you don't need it! And because one of the hypotheses (under which this technique applies) is not verified in this case.
Here, in layman terms, the ratio of functions, and the ratio of derivatives as well, does not have a clear enough limit. The de l'Hospital rule is more than often misused: a lot of people believe that if one cannot compute the limit of a ratio of functions, it is easier to compute easily the limit of the ratio of their derivatives. Possibly because derivatives sometimes look simpler, as for polynomials.
However, this is not true in general. Derivatives are rarely more continuous than original functions (my fundamental anti-theorem of analysis).
And thus said, remember that the purpose of exercises is to train your mathematical skills, not to get the result. The teacher knows it already (hopefully). Using de L'Hôpital's rule is sometimes overkill, with which you don't learn what is going on with your functions. It is more efficient, and sounder, to try first simpler techniques, such as factorization of leading terms (here $x$), transformations (logarithms), etc.
Now let us go to the point.
De L'Hôpital's rule states that: if $f$ and $g$ are functions that are differentiable on some (small enough) open interval $I$ (except possibly at a point $x_0$ contained in $I$), if $$lim_{xto x_0}f(x)=lim_{xto x_0}g(x)=0 ;mathrm{ or }; pminfty,$$ if $g'(x)ne 0$ for all $x$ in $I$ with $x ne x_0$, and $lim_{xto x_0}frac{f'(x)}{g'(x)}$ exists, then:
$$lim_{xto x_0}frac{f(x)}{g(x)} = lim_{xto x_0}frac{f'(x)}{g'(x)},.$$
The most classical "counter-example" is when functions are constant: $f(x)=c$ and $g(x)=1$. The derivative of $g(x)$ vanishes on any open interval, while $f/g = c$.
The factorization proposed by @Nick Peterson typically avoids to resort to this overkill rule when it is not necessary (especially when the indeterminacy can be lifted easily).
De L'Hôpital's rule looks magic, and as for every magic, it shall be used wisely with parsimony (unless it unleashes terrible powers).
Here, tracks are legion. One very simple is that $x$ grows as... $x$, and $sin(x)$ is bounded between $-1$ and $1$. So for a big engough $x$, $x-1le x+sin(x) le x+1$, and you know that$ frac{x}{x+1} $ and $frac{x}{x-1}$ clearly tends to $1$.
answered Mar 24 '16 at 6:28
Laurent DuvalLaurent Duval
5,37311240
$endgroup$
add a comment |
$begingroup$
others already said that l'Hopital requires existence of the limit of the ratio of the derivatives;
However in addition, with a solid understanding of limit definition is still possible to prove solution applying De l'Hopital, but not to that function, think about this:
$$lim_{x to +infty} frac{x}{x+1} leq lim_{x to +infty} frac{x}{x-sin(x)} leq lim_{x to +infty} frac{x}{x-1}$$
condensed considering also $-infty$ with
$$lim_{x to infty} frac{x}{x+sig(x)} leq lim_{x to infty} frac{x}{x+sin(x)} leq lim_{x to infty} frac{x}{x-sig(x)}$$
where
$$sig(x)=left{
begin{matrix}
0 & x=0\
frac{|x|}x & xne 0
end{matrix}
right.$$
prove the above while apply l'Hopital to
$$lim_{x to infty} frac{x}{xpm 1}$$
the squeezing inequities are true after a certain G, formally $exists G / forall xinRe,|x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
applying the limit definition to $x over x+sin(x)$ the starting point M selecting all x>M has to be greater or equal than G (simply require $Mgeq G$), in this case M=G is great enough to say that the limit is the same 1.
More formally (I actually didn't find an online pointable suitable formal definition of $lim_{xtoinfty}$, so I'm making it up)
$$lim_{x to infty} f(x) = rin {Re, -infty, +infty, NaN} / \
exists r in Re : forall epsilon in Re, epsilon>0: exists M in Re : forall x in Re, |x| > M : |f(x)-r|<epsilon \
lor r=infty, omissis \
lor r=+infty, omissis \
lor r=-infty, omissis \
lor r=NaN, omissis. $$
(r as abbreviation of response, NaN (not a number) is when the limit doesn't exists and $lor$ is in this case a shortcut or).
think of names
$f(x)=frac{x}{x+sin(x)}$
$g(x)=frac{x}{x pm 1}$, and when the definition of limit is used with g(x) the lower bound M is called G
from the evident property
$exists G' in Re^+ | forall x in Re, |x|>G' : x-1 leq x+sin(x) leq x+1$
$Rightarrow exists G in Re^+ | forall x in Re, |x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
$$lim_{x to infty} frac{x}{xpm 1} underleftarrow{=(?H)= lim_{x to +infty} frac{frac{d}{dx} x}{frac{d}{dx}(x pm 1)} = lim_{x to +infty} frac{1}{1 pm 0}=1}$$
the existence of this limits (they are two, due to $pm$) ensures that
$forall epsilon in Re, epsilon>0: exists G in Re : forall x in Re, |x| > G : |g(x)-r|<epsilon$
Choosing $M geq G$ ($M$ is the lower bound in the definition of limit for $f(x)$)
$$ Rightarrow
lim_{x to infty} f(x)=1$$
answered Mar 24 '16 at 19:03
Marco MunariMarco Munari
1094
$endgroup$
add a comment |
$begingroup$
There is another useful rule, which I don't seem to have seen written down explicitly:
Let $f, g, r$ and $s$ be functions such that $gtoinfty$ and $r, s$ are bounded.
Then the limit of $dfrac{f}{g}$ and the limit of $dfrac{f + r}{g + s}$ gives the same result.
Applied here, since $sin x$ is bounded, the limit is the same as the limit of $dfrac{x}{x}$.
edited Oct 28 '16 at 18:24
Bumblebee
9,72912551
answered Mar 25 '16 at 21:16
gnasher729gnasher729
6,1401128
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1710786%2fwhy-does-lhopitals-rule-fail-in-this-case%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule. What the rules says is IF the limit of $f'$ over $g'$ exists then the limit of $f$ over $g$ also exists and the two limits are the same. It doesn't say anything if the limit of $f'$ over $g'$ doesn't exist.
answered Mar 23 '16 at 21:17
Barry CipraBarry Cipra
60.5k655129
$endgroup$
67
$begingroup$
Not just a common error, a VERY common error. And I am from now on going to use the OP's example in my next calc batch!
$endgroup$
– imranfat
Mar 23 '16 at 21:18
6
$begingroup$
What's even funnier is examples where you apply l'Hopital twice and get back where you started :-)
$endgroup$
– gnasher729
Mar 25 '16 at 21:11
add a comment |
$begingroup$
Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule. What the rules says is IF the limit of $f'$ over $g'$ exists then the limit of $f$ over $g$ also exists and the two limits are the same. It doesn't say anything if the limit of $f'$ over $g'$ doesn't exist.
answered Mar 23 '16 at 21:17
Barry CipraBarry Cipra
60.5k655129
$endgroup$
67
$begingroup$
Not just a common error, a VERY common error. And I am from now on going to use the OP's example in my next calc batch!
$endgroup$
– imranfat
Mar 23 '16 at 21:18
6
$begingroup$
What's even funnier is examples where you apply l'Hopital twice and get back where you started :-)
$endgroup$
– gnasher729
Mar 25 '16 at 21:11
add a comment |
$begingroup$
Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule. What the rules says is IF the limit of $f'$ over $g'$ exists then the limit of $f$ over $g$ also exists and the two limits are the same. It doesn't say anything if the limit of $f'$ over $g'$ doesn't exist.
answered Mar 23 '16 at 21:17
Barry CipraBarry Cipra
60.5k655129
$endgroup$
Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule. What the rules says is IF the limit of $f'$ over $g'$ exists then the limit of $f$ over $g$ also exists and the two limits are the same. It doesn't say anything if the limit of $f'$ over $g'$ doesn't exist.
answered Mar 23 '16 at 21:17
Barry CipraBarry Cipra
60.5k655129
answered Mar 23 '16 at 21:17
Barry CipraBarry Cipra
60.5k655129
answered Mar 23 '16 at 21:17
Barry CipraBarry Cipra
60.5k655129
answered Mar 23 '16 at 21:17
Barry CipraBarry Cipra
60.5k655129
60.5k655129
67
$begingroup$
Not just a common error, a VERY common error. And I am from now on going to use the OP's example in my next calc batch!
$endgroup$
– imranfat
Mar 23 '16 at 21:18
6
$begingroup$
What's even funnier is examples where you apply l'Hopital twice and get back where you started :-)
$endgroup$
– gnasher729
Mar 25 '16 at 21:11
add a comment |
67
$begingroup$
Not just a common error, a VERY common error. And I am from now on going to use the OP's example in my next calc batch!
$endgroup$
– imranfat
Mar 23 '16 at 21:18
6
$begingroup$
What's even funnier is examples where you apply l'Hopital twice and get back where you started :-)
$endgroup$
– gnasher729
Mar 25 '16 at 21:11
67
67
$begingroup$
Not just a common error, a VERY common error. And I am from now on going to use the OP's example in my next calc batch!
$endgroup$
– imranfat
Mar 23 '16 at 21:18
$begingroup$
Not just a common error, a VERY common error. And I am from now on going to use the OP's example in my next calc batch!
$endgroup$
– imranfat
Mar 23 '16 at 21:18
6
6
$begingroup$
What's even funnier is examples where you apply l'Hopital twice and get back where you started :-)
$endgroup$
– gnasher729
Mar 25 '16 at 21:11
$begingroup$
What's even funnier is examples where you apply l'Hopital twice and get back where you started :-)
$endgroup$
– gnasher729
Mar 25 '16 at 21:11
add a comment |
$begingroup$
L'Hopital's rule only tells you that if the modified limit exists and has value $L$, then the original limit also exists and has value $L$. It doesn't tell you that the converse holds.
So, the fact that the modified limit doesn't exist gives you no information about the original limit. So, you need a different method.
Consider something more direct: can you compute
$$
lim_{xtoinfty}frac{x}{x+sin x}=lim_{xtoinfty}frac{1}{1+frac{sin x}{x}}?
$$
answered Mar 23 '16 at 21:12
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Or 1 - sin x / (x + sin x).
$endgroup$
– gnasher729
Mar 24 '16 at 22:16
2
$begingroup$
or squeeze theorem using -1 <= sin x <= 1.
$endgroup$
– djechlin
Mar 25 '16 at 18:57
add a comment |
$begingroup$
L'Hopital's rule only tells you that if the modified limit exists and has value $L$, then the original limit also exists and has value $L$. It doesn't tell you that the converse holds.
So, the fact that the modified limit doesn't exist gives you no information about the original limit. So, you need a different method.
Consider something more direct: can you compute
$$
lim_{xtoinfty}frac{x}{x+sin x}=lim_{xtoinfty}frac{1}{1+frac{sin x}{x}}?
$$
answered Mar 23 '16 at 21:12
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Or 1 - sin x / (x + sin x).
$endgroup$
– gnasher729
Mar 24 '16 at 22:16
2
$begingroup$
or squeeze theorem using -1 <= sin x <= 1.
$endgroup$
– djechlin
Mar 25 '16 at 18:57
add a comment |
$begingroup$
L'Hopital's rule only tells you that if the modified limit exists and has value $L$, then the original limit also exists and has value $L$. It doesn't tell you that the converse holds.
So, the fact that the modified limit doesn't exist gives you no information about the original limit. So, you need a different method.
Consider something more direct: can you compute
$$
lim_{xtoinfty}frac{x}{x+sin x}=lim_{xtoinfty}frac{1}{1+frac{sin x}{x}}?
$$
answered Mar 23 '16 at 21:12
Nick PetersonNick Peterson
26.8k23962
$endgroup$
L'Hopital's rule only tells you that if the modified limit exists and has value $L$, then the original limit also exists and has value $L$. It doesn't tell you that the converse holds.
So, the fact that the modified limit doesn't exist gives you no information about the original limit. So, you need a different method.
Consider something more direct: can you compute
$$
lim_{xtoinfty}frac{x}{x+sin x}=lim_{xtoinfty}frac{1}{1+frac{sin x}{x}}?
$$
answered Mar 23 '16 at 21:12
Nick PetersonNick Peterson
26.8k23962
answered Mar 23 '16 at 21:12
Nick PetersonNick Peterson
26.8k23962
answered Mar 23 '16 at 21:12
Nick PetersonNick Peterson
26.8k23962
answered Mar 23 '16 at 21:12
Nick PetersonNick Peterson
26.8k23962
26.8k23962
$begingroup$
Or 1 - sin x / (x + sin x).
$endgroup$
– gnasher729
Mar 24 '16 at 22:16
2
$begingroup$
or squeeze theorem using -1 <= sin x <= 1.
$endgroup$
– djechlin
Mar 25 '16 at 18:57
add a comment |
$begingroup$
Or 1 - sin x / (x + sin x).
$endgroup$
– gnasher729
Mar 24 '16 at 22:16
2
$begingroup$
or squeeze theorem using -1 <= sin x <= 1.
$endgroup$
– djechlin
Mar 25 '16 at 18:57
$begingroup$
Or 1 - sin x / (x + sin x).
$endgroup$
– gnasher729
Mar 24 '16 at 22:16
$begingroup$
Or 1 - sin x / (x + sin x).
$endgroup$
– gnasher729
Mar 24 '16 at 22:16
2
2
$begingroup$
or squeeze theorem using -1 <= sin x <= 1.
$endgroup$
– djechlin
Mar 25 '16 at 18:57
$begingroup$
or squeeze theorem using -1 <= sin x <= 1.
$endgroup$
– djechlin
Mar 25 '16 at 18:57
add a comment |
$begingroup$
Because you don't need it! And because one of the hypotheses (under which this technique applies) is not verified in this case.
Here, in layman terms, the ratio of functions, and the ratio of derivatives as well, does not have a clear enough limit. The de l'Hospital rule is more than often misused: a lot of people believe that if one cannot compute the limit of a ratio of functions, it is easier to compute easily the limit of the ratio of their derivatives. Possibly because derivatives sometimes look simpler, as for polynomials.
However, this is not true in general. Derivatives are rarely more continuous than original functions (my fundamental anti-theorem of analysis).
And thus said, remember that the purpose of exercises is to train your mathematical skills, not to get the result. The teacher knows it already (hopefully). Using de L'Hôpital's rule is sometimes overkill, with which you don't learn what is going on with your functions. It is more efficient, and sounder, to try first simpler techniques, such as factorization of leading terms (here $x$), transformations (logarithms), etc.
Now let us go to the point.
De L'Hôpital's rule states that: if $f$ and $g$ are functions that are differentiable on some (small enough) open interval $I$ (except possibly at a point $x_0$ contained in $I$), if $$lim_{xto x_0}f(x)=lim_{xto x_0}g(x)=0 ;mathrm{ or }; pminfty,$$ if $g'(x)ne 0$ for all $x$ in $I$ with $x ne x_0$, and $lim_{xto x_0}frac{f'(x)}{g'(x)}$ exists, then:
$$lim_{xto x_0}frac{f(x)}{g(x)} = lim_{xto x_0}frac{f'(x)}{g'(x)},.$$
The most classical "counter-example" is when functions are constant: $f(x)=c$ and $g(x)=1$. The derivative of $g(x)$ vanishes on any open interval, while $f/g = c$.
The factorization proposed by @Nick Peterson typically avoids to resort to this overkill rule when it is not necessary (especially when the indeterminacy can be lifted easily).
De L'Hôpital's rule looks magic, and as for every magic, it shall be used wisely with parsimony (unless it unleashes terrible powers).
Here, tracks are legion. One very simple is that $x$ grows as... $x$, and $sin(x)$ is bounded between $-1$ and $1$. So for a big engough $x$, $x-1le x+sin(x) le x+1$, and you know that$ frac{x}{x+1} $ and $frac{x}{x-1}$ clearly tends to $1$.
answered Mar 24 '16 at 6:28
Laurent DuvalLaurent Duval
5,37311240
$endgroup$
add a comment |
$begingroup$
Because you don't need it! And because one of the hypotheses (under which this technique applies) is not verified in this case.
Here, in layman terms, the ratio of functions, and the ratio of derivatives as well, does not have a clear enough limit. The de l'Hospital rule is more than often misused: a lot of people believe that if one cannot compute the limit of a ratio of functions, it is easier to compute easily the limit of the ratio of their derivatives. Possibly because derivatives sometimes look simpler, as for polynomials.
However, this is not true in general. Derivatives are rarely more continuous than original functions (my fundamental anti-theorem of analysis).
And thus said, remember that the purpose of exercises is to train your mathematical skills, not to get the result. The teacher knows it already (hopefully). Using de L'Hôpital's rule is sometimes overkill, with which you don't learn what is going on with your functions. It is more efficient, and sounder, to try first simpler techniques, such as factorization of leading terms (here $x$), transformations (logarithms), etc.
Now let us go to the point.
De L'Hôpital's rule states that: if $f$ and $g$ are functions that are differentiable on some (small enough) open interval $I$ (except possibly at a point $x_0$ contained in $I$), if $$lim_{xto x_0}f(x)=lim_{xto x_0}g(x)=0 ;mathrm{ or }; pminfty,$$ if $g'(x)ne 0$ for all $x$ in $I$ with $x ne x_0$, and $lim_{xto x_0}frac{f'(x)}{g'(x)}$ exists, then:
$$lim_{xto x_0}frac{f(x)}{g(x)} = lim_{xto x_0}frac{f'(x)}{g'(x)},.$$
The most classical "counter-example" is when functions are constant: $f(x)=c$ and $g(x)=1$. The derivative of $g(x)$ vanishes on any open interval, while $f/g = c$.
The factorization proposed by @Nick Peterson typically avoids to resort to this overkill rule when it is not necessary (especially when the indeterminacy can be lifted easily).
De L'Hôpital's rule looks magic, and as for every magic, it shall be used wisely with parsimony (unless it unleashes terrible powers).
Here, tracks are legion. One very simple is that $x$ grows as... $x$, and $sin(x)$ is bounded between $-1$ and $1$. So for a big engough $x$, $x-1le x+sin(x) le x+1$, and you know that$ frac{x}{x+1} $ and $frac{x}{x-1}$ clearly tends to $1$.
answered Mar 24 '16 at 6:28
Laurent DuvalLaurent Duval
5,37311240
$endgroup$
add a comment |
$begingroup$
Because you don't need it! And because one of the hypotheses (under which this technique applies) is not verified in this case.
Here, in layman terms, the ratio of functions, and the ratio of derivatives as well, does not have a clear enough limit. The de l'Hospital rule is more than often misused: a lot of people believe that if one cannot compute the limit of a ratio of functions, it is easier to compute easily the limit of the ratio of their derivatives. Possibly because derivatives sometimes look simpler, as for polynomials.
However, this is not true in general. Derivatives are rarely more continuous than original functions (my fundamental anti-theorem of analysis).
And thus said, remember that the purpose of exercises is to train your mathematical skills, not to get the result. The teacher knows it already (hopefully). Using de L'Hôpital's rule is sometimes overkill, with which you don't learn what is going on with your functions. It is more efficient, and sounder, to try first simpler techniques, such as factorization of leading terms (here $x$), transformations (logarithms), etc.
Now let us go to the point.
De L'Hôpital's rule states that: if $f$ and $g$ are functions that are differentiable on some (small enough) open interval $I$ (except possibly at a point $x_0$ contained in $I$), if $$lim_{xto x_0}f(x)=lim_{xto x_0}g(x)=0 ;mathrm{ or }; pminfty,$$ if $g'(x)ne 0$ for all $x$ in $I$ with $x ne x_0$, and $lim_{xto x_0}frac{f'(x)}{g'(x)}$ exists, then:
$$lim_{xto x_0}frac{f(x)}{g(x)} = lim_{xto x_0}frac{f'(x)}{g'(x)},.$$
The most classical "counter-example" is when functions are constant: $f(x)=c$ and $g(x)=1$. The derivative of $g(x)$ vanishes on any open interval, while $f/g = c$.
The factorization proposed by @Nick Peterson typically avoids to resort to this overkill rule when it is not necessary (especially when the indeterminacy can be lifted easily).
De L'Hôpital's rule looks magic, and as for every magic, it shall be used wisely with parsimony (unless it unleashes terrible powers).
Here, tracks are legion. One very simple is that $x$ grows as... $x$, and $sin(x)$ is bounded between $-1$ and $1$. So for a big engough $x$, $x-1le x+sin(x) le x+1$, and you know that$ frac{x}{x+1} $ and $frac{x}{x-1}$ clearly tends to $1$.
answered Mar 24 '16 at 6:28
Laurent DuvalLaurent Duval
5,37311240
$endgroup$
Because you don't need it! And because one of the hypotheses (under which this technique applies) is not verified in this case.
Here, in layman terms, the ratio of functions, and the ratio of derivatives as well, does not have a clear enough limit. The de l'Hospital rule is more than often misused: a lot of people believe that if one cannot compute the limit of a ratio of functions, it is easier to compute easily the limit of the ratio of their derivatives. Possibly because derivatives sometimes look simpler, as for polynomials.
However, this is not true in general. Derivatives are rarely more continuous than original functions (my fundamental anti-theorem of analysis).
And thus said, remember that the purpose of exercises is to train your mathematical skills, not to get the result. The teacher knows it already (hopefully). Using de L'Hôpital's rule is sometimes overkill, with which you don't learn what is going on with your functions. It is more efficient, and sounder, to try first simpler techniques, such as factorization of leading terms (here $x$), transformations (logarithms), etc.
Now let us go to the point.
De L'Hôpital's rule states that: if $f$ and $g$ are functions that are differentiable on some (small enough) open interval $I$ (except possibly at a point $x_0$ contained in $I$), if $$lim_{xto x_0}f(x)=lim_{xto x_0}g(x)=0 ;mathrm{ or }; pminfty,$$ if $g'(x)ne 0$ for all $x$ in $I$ with $x ne x_0$, and $lim_{xto x_0}frac{f'(x)}{g'(x)}$ exists, then:
$$lim_{xto x_0}frac{f(x)}{g(x)} = lim_{xto x_0}frac{f'(x)}{g'(x)},.$$
The most classical "counter-example" is when functions are constant: $f(x)=c$ and $g(x)=1$. The derivative of $g(x)$ vanishes on any open interval, while $f/g = c$.
The factorization proposed by @Nick Peterson typically avoids to resort to this overkill rule when it is not necessary (especially when the indeterminacy can be lifted easily).
De L'Hôpital's rule looks magic, and as for every magic, it shall be used wisely with parsimony (unless it unleashes terrible powers).
Here, tracks are legion. One very simple is that $x$ grows as... $x$, and $sin(x)$ is bounded between $-1$ and $1$. So for a big engough $x$, $x-1le x+sin(x) le x+1$, and you know that$ frac{x}{x+1} $ and $frac{x}{x-1}$ clearly tends to $1$.
answered Mar 24 '16 at 6:28
Laurent DuvalLaurent Duval
5,37311240
edited Jan 29 at 21:29
answered Mar 24 '16 at 6:28
Laurent DuvalLaurent Duval
5,37311240
answered Mar 24 '16 at 6:28
Laurent DuvalLaurent Duval
5,37311240
answered Mar 24 '16 at 6:28
Laurent DuvalLaurent Duval
5,37311240
5,37311240
add a comment |
add a comment |
$begingroup$
others already said that l'Hopital requires existence of the limit of the ratio of the derivatives;
However in addition, with a solid understanding of limit definition is still possible to prove solution applying De l'Hopital, but not to that function, think about this:
$$lim_{x to +infty} frac{x}{x+1} leq lim_{x to +infty} frac{x}{x-sin(x)} leq lim_{x to +infty} frac{x}{x-1}$$
condensed considering also $-infty$ with
$$lim_{x to infty} frac{x}{x+sig(x)} leq lim_{x to infty} frac{x}{x+sin(x)} leq lim_{x to infty} frac{x}{x-sig(x)}$$
where
$$sig(x)=left{
begin{matrix}
0 & x=0\
frac{|x|}x & xne 0
end{matrix}
right.$$
prove the above while apply l'Hopital to
$$lim_{x to infty} frac{x}{xpm 1}$$
the squeezing inequities are true after a certain G, formally $exists G / forall xinRe,|x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
applying the limit definition to $x over x+sin(x)$ the starting point M selecting all x>M has to be greater or equal than G (simply require $Mgeq G$), in this case M=G is great enough to say that the limit is the same 1.
More formally (I actually didn't find an online pointable suitable formal definition of $lim_{xtoinfty}$, so I'm making it up)
$$lim_{x to infty} f(x) = rin {Re, -infty, +infty, NaN} / \
exists r in Re : forall epsilon in Re, epsilon>0: exists M in Re : forall x in Re, |x| > M : |f(x)-r|<epsilon \
lor r=infty, omissis \
lor r=+infty, omissis \
lor r=-infty, omissis \
lor r=NaN, omissis. $$
(r as abbreviation of response, NaN (not a number) is when the limit doesn't exists and $lor$ is in this case a shortcut or).
think of names
$f(x)=frac{x}{x+sin(x)}$
$g(x)=frac{x}{x pm 1}$, and when the definition of limit is used with g(x) the lower bound M is called G
from the evident property
$exists G' in Re^+ | forall x in Re, |x|>G' : x-1 leq x+sin(x) leq x+1$
$Rightarrow exists G in Re^+ | forall x in Re, |x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
$$lim_{x to infty} frac{x}{xpm 1} underleftarrow{=(?H)= lim_{x to +infty} frac{frac{d}{dx} x}{frac{d}{dx}(x pm 1)} = lim_{x to +infty} frac{1}{1 pm 0}=1}$$
the existence of this limits (they are two, due to $pm$) ensures that
$forall epsilon in Re, epsilon>0: exists G in Re : forall x in Re, |x| > G : |g(x)-r|<epsilon$
Choosing $M geq G$ ($M$ is the lower bound in the definition of limit for $f(x)$)
$$ Rightarrow
lim_{x to infty} f(x)=1$$
answered Mar 24 '16 at 19:03
Marco MunariMarco Munari
1094
$endgroup$
add a comment |
$begingroup$
others already said that l'Hopital requires existence of the limit of the ratio of the derivatives;
However in addition, with a solid understanding of limit definition is still possible to prove solution applying De l'Hopital, but not to that function, think about this:
$$lim_{x to +infty} frac{x}{x+1} leq lim_{x to +infty} frac{x}{x-sin(x)} leq lim_{x to +infty} frac{x}{x-1}$$
condensed considering also $-infty$ with
$$lim_{x to infty} frac{x}{x+sig(x)} leq lim_{x to infty} frac{x}{x+sin(x)} leq lim_{x to infty} frac{x}{x-sig(x)}$$
where
$$sig(x)=left{
begin{matrix}
0 & x=0\
frac{|x|}x & xne 0
end{matrix}
right.$$
prove the above while apply l'Hopital to
$$lim_{x to infty} frac{x}{xpm 1}$$
the squeezing inequities are true after a certain G, formally $exists G / forall xinRe,|x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
applying the limit definition to $x over x+sin(x)$ the starting point M selecting all x>M has to be greater or equal than G (simply require $Mgeq G$), in this case M=G is great enough to say that the limit is the same 1.
More formally (I actually didn't find an online pointable suitable formal definition of $lim_{xtoinfty}$, so I'm making it up)
$$lim_{x to infty} f(x) = rin {Re, -infty, +infty, NaN} / \
exists r in Re : forall epsilon in Re, epsilon>0: exists M in Re : forall x in Re, |x| > M : |f(x)-r|<epsilon \
lor r=infty, omissis \
lor r=+infty, omissis \
lor r=-infty, omissis \
lor r=NaN, omissis. $$
(r as abbreviation of response, NaN (not a number) is when the limit doesn't exists and $lor$ is in this case a shortcut or).
think of names
$f(x)=frac{x}{x+sin(x)}$
$g(x)=frac{x}{x pm 1}$, and when the definition of limit is used with g(x) the lower bound M is called G
from the evident property
$exists G' in Re^+ | forall x in Re, |x|>G' : x-1 leq x+sin(x) leq x+1$
$Rightarrow exists G in Re^+ | forall x in Re, |x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
$$lim_{x to infty} frac{x}{xpm 1} underleftarrow{=(?H)= lim_{x to +infty} frac{frac{d}{dx} x}{frac{d}{dx}(x pm 1)} = lim_{x to +infty} frac{1}{1 pm 0}=1}$$
the existence of this limits (they are two, due to $pm$) ensures that
$forall epsilon in Re, epsilon>0: exists G in Re : forall x in Re, |x| > G : |g(x)-r|<epsilon$
Choosing $M geq G$ ($M$ is the lower bound in the definition of limit for $f(x)$)
$$ Rightarrow
lim_{x to infty} f(x)=1$$
answered Mar 24 '16 at 19:03
Marco MunariMarco Munari
1094
$endgroup$
add a comment |
$begingroup$
others already said that l'Hopital requires existence of the limit of the ratio of the derivatives;
However in addition, with a solid understanding of limit definition is still possible to prove solution applying De l'Hopital, but not to that function, think about this:
$$lim_{x to +infty} frac{x}{x+1} leq lim_{x to +infty} frac{x}{x-sin(x)} leq lim_{x to +infty} frac{x}{x-1}$$
condensed considering also $-infty$ with
$$lim_{x to infty} frac{x}{x+sig(x)} leq lim_{x to infty} frac{x}{x+sin(x)} leq lim_{x to infty} frac{x}{x-sig(x)}$$
where
$$sig(x)=left{
begin{matrix}
0 & x=0\
frac{|x|}x & xne 0
end{matrix}
right.$$
prove the above while apply l'Hopital to
$$lim_{x to infty} frac{x}{xpm 1}$$
the squeezing inequities are true after a certain G, formally $exists G / forall xinRe,|x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
applying the limit definition to $x over x+sin(x)$ the starting point M selecting all x>M has to be greater or equal than G (simply require $Mgeq G$), in this case M=G is great enough to say that the limit is the same 1.
More formally (I actually didn't find an online pointable suitable formal definition of $lim_{xtoinfty}$, so I'm making it up)
$$lim_{x to infty} f(x) = rin {Re, -infty, +infty, NaN} / \
exists r in Re : forall epsilon in Re, epsilon>0: exists M in Re : forall x in Re, |x| > M : |f(x)-r|<epsilon \
lor r=infty, omissis \
lor r=+infty, omissis \
lor r=-infty, omissis \
lor r=NaN, omissis. $$
(r as abbreviation of response, NaN (not a number) is when the limit doesn't exists and $lor$ is in this case a shortcut or).
think of names
$f(x)=frac{x}{x+sin(x)}$
$g(x)=frac{x}{x pm 1}$, and when the definition of limit is used with g(x) the lower bound M is called G
from the evident property
$exists G' in Re^+ | forall x in Re, |x|>G' : x-1 leq x+sin(x) leq x+1$
$Rightarrow exists G in Re^+ | forall x in Re, |x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
$$lim_{x to infty} frac{x}{xpm 1} underleftarrow{=(?H)= lim_{x to +infty} frac{frac{d}{dx} x}{frac{d}{dx}(x pm 1)} = lim_{x to +infty} frac{1}{1 pm 0}=1}$$
the existence of this limits (they are two, due to $pm$) ensures that
$forall epsilon in Re, epsilon>0: exists G in Re : forall x in Re, |x| > G : |g(x)-r|<epsilon$
Choosing $M geq G$ ($M$ is the lower bound in the definition of limit for $f(x)$)
$$ Rightarrow
lim_{x to infty} f(x)=1$$
answered Mar 24 '16 at 19:03
Marco MunariMarco Munari
1094
$endgroup$
others already said that l'Hopital requires existence of the limit of the ratio of the derivatives;
However in addition, with a solid understanding of limit definition is still possible to prove solution applying De l'Hopital, but not to that function, think about this:
$$lim_{x to +infty} frac{x}{x+1} leq lim_{x to +infty} frac{x}{x-sin(x)} leq lim_{x to +infty} frac{x}{x-1}$$
condensed considering also $-infty$ with
$$lim_{x to infty} frac{x}{x+sig(x)} leq lim_{x to infty} frac{x}{x+sin(x)} leq lim_{x to infty} frac{x}{x-sig(x)}$$
where
$$sig(x)=left{
begin{matrix}
0 & x=0\
frac{|x|}x & xne 0
end{matrix}
right.$$
prove the above while apply l'Hopital to
$$lim_{x to infty} frac{x}{xpm 1}$$
the squeezing inequities are true after a certain G, formally $exists G / forall xinRe,|x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
applying the limit definition to $x over x+sin(x)$ the starting point M selecting all x>M has to be greater or equal than G (simply require $Mgeq G$), in this case M=G is great enough to say that the limit is the same 1.
More formally (I actually didn't find an online pointable suitable formal definition of $lim_{xtoinfty}$, so I'm making it up)
$$lim_{x to infty} f(x) = rin {Re, -infty, +infty, NaN} / \
exists r in Re : forall epsilon in Re, epsilon>0: exists M in Re : forall x in Re, |x| > M : |f(x)-r|<epsilon \
lor r=infty, omissis \
lor r=+infty, omissis \
lor r=-infty, omissis \
lor r=NaN, omissis. $$
(r as abbreviation of response, NaN (not a number) is when the limit doesn't exists and $lor$ is in this case a shortcut or).
think of names
$f(x)=frac{x}{x+sin(x)}$
$g(x)=frac{x}{x pm 1}$, and when the definition of limit is used with g(x) the lower bound M is called G
from the evident property
$exists G' in Re^+ | forall x in Re, |x|>G' : x-1 leq x+sin(x) leq x+1$
$Rightarrow exists G in Re^+ | forall x in Re, |x|>G : frac{x}{x+sig(x)} leq frac{x}{x+sin(x)} leq frac{x}{x-sig(x)}$
$$lim_{x to infty} frac{x}{xpm 1} underleftarrow{=(?H)= lim_{x to +infty} frac{frac{d}{dx} x}{frac{d}{dx}(x pm 1)} = lim_{x to +infty} frac{1}{1 pm 0}=1}$$
the existence of this limits (they are two, due to $pm$) ensures that
$forall epsilon in Re, epsilon>0: exists G in Re : forall x in Re, |x| > G : |g(x)-r|<epsilon$
Choosing $M geq G$ ($M$ is the lower bound in the definition of limit for $f(x)$)
$$ Rightarrow
lim_{x to infty} f(x)=1$$
answered Mar 24 '16 at 19:03
Marco MunariMarco Munari
1094
edited Mar 25 '16 at 5:48
answered Mar 24 '16 at 19:03
Marco MunariMarco Munari
1094
answered Mar 24 '16 at 19:03
Marco MunariMarco Munari
1094
answered Mar 24 '16 at 19:03
Marco MunariMarco Munari
1094
1094
add a comment |
add a comment |
$begingroup$
There is another useful rule, which I don't seem to have seen written down explicitly:
Let $f, g, r$ and $s$ be functions such that $gtoinfty$ and $r, s$ are bounded.
Then the limit of $dfrac{f}{g}$ and the limit of $dfrac{f + r}{g + s}$ gives the same result.
Applied here, since $sin x$ is bounded, the limit is the same as the limit of $dfrac{x}{x}$.
edited Oct 28 '16 at 18:24
Bumblebee
9,72912551
answered Mar 25 '16 at 21:16
gnasher729gnasher729
6,1401128
$endgroup$
add a comment |
$begingroup$
There is another useful rule, which I don't seem to have seen written down explicitly:
Let $f, g, r$ and $s$ be functions such that $gtoinfty$ and $r, s$ are bounded.
Then the limit of $dfrac{f}{g}$ and the limit of $dfrac{f + r}{g + s}$ gives the same result.
Applied here, since $sin x$ is bounded, the limit is the same as the limit of $dfrac{x}{x}$.
edited Oct 28 '16 at 18:24
Bumblebee
9,72912551
answered Mar 25 '16 at 21:16
gnasher729gnasher729
6,1401128
$endgroup$
add a comment |
$begingroup$
There is another useful rule, which I don't seem to have seen written down explicitly:
Let $f, g, r$ and $s$ be functions such that $gtoinfty$ and $r, s$ are bounded.
Then the limit of $dfrac{f}{g}$ and the limit of $dfrac{f + r}{g + s}$ gives the same result.
Applied here, since $sin x$ is bounded, the limit is the same as the limit of $dfrac{x}{x}$.
edited Oct 28 '16 at 18:24
Bumblebee
9,72912551
answered Mar 25 '16 at 21:16
gnasher729gnasher729
6,1401128
$endgroup$
There is another useful rule, which I don't seem to have seen written down explicitly:
Let $f, g, r$ and $s$ be functions such that $gtoinfty$ and $r, s$ are bounded.
Then the limit of $dfrac{f}{g}$ and the limit of $dfrac{f + r}{g + s}$ gives the same result.
Applied here, since $sin x$ is bounded, the limit is the same as the limit of $dfrac{x}{x}$.
edited Oct 28 '16 at 18:24
Bumblebee
9,72912551
answered Mar 25 '16 at 21:16
gnasher729gnasher729
6,1401128
edited Oct 28 '16 at 18:24
Bumblebee
9,72912551
edited Oct 28 '16 at 18:24
Bumblebee
9,72912551
edited Oct 28 '16 at 18:24
Bumblebee
9,72912551
9,72912551
answered Mar 25 '16 at 21:16
gnasher729gnasher729
6,1401128
answered Mar 25 '16 at 21:16
gnasher729gnasher729
6,1401128
answered Mar 25 '16 at 21:16
gnasher729gnasher729
6,1401128
6,1401128
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1710786%2fwhy-does-lhopitals-rule-fail-in-this-case%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
75
$begingroup$
If the limit of $f'/g'$ exists, then it is also the limit of $f/g$. Not the other way around.
$endgroup$
– user251257
Mar 23 '16 at 21:11
8
$begingroup$
One often forgets there are hypotheses to check before applying L'Hospital. One of these is that the ratio of the derivatives must exist (or still be indeterminate).
$endgroup$
– Bernard
Mar 23 '16 at 21:13
3
$begingroup$
A condition on the use of L'Hôpital in this context is that the derivative of the denominator must be non-zero on $(N, infty)$ for some $N$.
$endgroup$
– Brian Tung
Mar 23 '16 at 21:15
2
$begingroup$
This post might give you something to think about.
$endgroup$
– Hirshy
Mar 23 '16 at 21:23
6
$begingroup$
An excellent example for a Calculus Course.
$endgroup$
– dwarandae
Mar 24 '16 at 4:46