Mixing
How exactly do I go through proving a limit exists with $epsilon$-$delta$? For instance, assuming $lim...
$begingroup$
Recently, I've been looking back into my calculus textbooks and noticed that we haven't gone over the rigorous proving of the limit. I think I understand the definition of the limit however, I'm having trouble proving some of the properties of the limits.
For example, looking at the limit property of a product of a constant and a function.
Assuming $$ lim_{x to c} f(x) = L $$
Prove that
$$lim_{x to c} k f(x) = kL = klim_{x to c} f(x) $$
First I must define $epsilon > 0$ and $k neq 0$. I need to show that
$$forall epsilon, existsdelta$$ such that $$0<|x -c|<delta Rightarrow |kf(x) -kL|<epsilon$$
Since we defined that the limit of $f(x)$ when $x$ goes to $c$ is $L$ the following statement must be true.
$$forall epsilon, existsdelta$$ such that $$0<|x -c|<delta Rightarrow |f(x) -L|<epsilon$$
Now I'm kinda stuck and can't seem to figure out what I should do.
real-analysis limits
edited Jan 13 at 0:24
Blue
48.4k870154
asked Jan 13 at 0:02
Nick YarnNick Yarn
236
$endgroup$
add a comment |
$begingroup$
Recently, I've been looking back into my calculus textbooks and noticed that we haven't gone over the rigorous proving of the limit. I think I understand the definition of the limit however, I'm having trouble proving some of the properties of the limits.
For example, looking at the limit property of a product of a constant and a function.
Assuming $$ lim_{x to c} f(x) = L $$
Prove that
$$lim_{x to c} k f(x) = kL = klim_{x to c} f(x) $$
First I must define $epsilon > 0$ and $k neq 0$. I need to show that
$$forall epsilon, existsdelta$$ such that $$0<|x -c|<delta Rightarrow |kf(x) -kL|<epsilon$$
Since we defined that the limit of $f(x)$ when $x$ goes to $c$ is $L$ the following statement must be true.
$$forall epsilon, existsdelta$$ such that $$0<|x -c|<delta Rightarrow |f(x) -L|<epsilon$$
Now I'm kinda stuck and can't seem to figure out what I should do.
real-analysis limits
edited Jan 13 at 0:24
Blue
48.4k870154
asked Jan 13 at 0:02
Nick YarnNick Yarn
236
$endgroup$
add a comment |
$begingroup$
Recently, I've been looking back into my calculus textbooks and noticed that we haven't gone over the rigorous proving of the limit. I think I understand the definition of the limit however, I'm having trouble proving some of the properties of the limits.
For example, looking at the limit property of a product of a constant and a function.
Assuming $$ lim_{x to c} f(x) = L $$
Prove that
$$lim_{x to c} k f(x) = kL = klim_{x to c} f(x) $$
First I must define $epsilon > 0$ and $k neq 0$. I need to show that
$$forall epsilon, existsdelta$$ such that $$0<|x -c|<delta Rightarrow |kf(x) -kL|<epsilon$$
Since we defined that the limit of $f(x)$ when $x$ goes to $c$ is $L$ the following statement must be true.
$$forall epsilon, existsdelta$$ such that $$0<|x -c|<delta Rightarrow |f(x) -L|<epsilon$$
Now I'm kinda stuck and can't seem to figure out what I should do.
real-analysis limits
edited Jan 13 at 0:24
Blue
48.4k870154
asked Jan 13 at 0:02
Nick YarnNick Yarn
236
$endgroup$
Recently, I've been looking back into my calculus textbooks and noticed that we haven't gone over the rigorous proving of the limit. I think I understand the definition of the limit however, I'm having trouble proving some of the properties of the limits.
For example, looking at the limit property of a product of a constant and a function.
Assuming $$ lim_{x to c} f(x) = L $$
Prove that
$$lim_{x to c} k f(x) = kL = klim_{x to c} f(x) $$
First I must define $epsilon > 0$ and $k neq 0$. I need to show that
$$forall epsilon, existsdelta$$ such that $$0<|x -c|<delta Rightarrow |kf(x) -kL|<epsilon$$
Since we defined that the limit of $f(x)$ when $x$ goes to $c$ is $L$ the following statement must be true.
$$forall epsilon, existsdelta$$ such that $$0<|x -c|<delta Rightarrow |f(x) -L|<epsilon$$
Now I'm kinda stuck and can't seem to figure out what I should do.
real-analysis limits
real-analysis limits
edited Jan 13 at 0:24
Blue
48.4k870154
asked Jan 13 at 0:02
Nick YarnNick Yarn
236
edited Jan 13 at 0:24
Blue
48.4k870154
asked Jan 13 at 0:02
Nick YarnNick Yarn
236
edited Jan 13 at 0:24
Blue
48.4k870154
edited Jan 13 at 0:24
Blue
48.4k870154
edited Jan 13 at 0:24
Blue
48.4k870154
48.4k870154
asked Jan 13 at 0:02
Nick YarnNick Yarn
236
asked Jan 13 at 0:02
Nick YarnNick Yarn
236
asked Jan 13 at 0:02
Nick YarnNick Yarn
236
236
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I think one thing that eludes must students is that although the $delta$s and $epsilon$s are technically constants, the choice of $delta$ is dependent upon both the $epsilon$ and the function.
So we know that because God told us $limlimits_{xto c} f(x) = L$. That we know for any $epsilon > 0$ there exists a $delta_{epsilon, f}$ so the whenever $|x - c| < delta_{epsilon, f}$ then it will occur that $|f(x) -c | < epsilon$.
So why did I write this number as $delta_{epsilon, f}$ instead of just $delta$ as must texts do? I didn't have to but I did because I wanted to emphasize this particular value is determined by and "specifically tailored" for that particular $epsilon$.
So we need to prove that for any $epsilon$ there is a $delta_{epsilon, 5f}$ so that whenever $|x-c|<delta_{epsilon, 5f}$ we will have $|5f(x) - 5L| < delta_{epsilon, 5f}$. How do we find that $delta_{epsilon, 5f}$?
We $|5f(x) - 5L| = 5|f(x) - L|$ so $|5f(x) - 5L|< epsilon iff |f(x) - L| < frac 15 epsilon$. And we know that $limlimits_{xto c} f(x) = L$ so for any $epsilon > 0$ we have $delta_{epsilon f}$ that does what we want.
Well, $frac 15 epsilon > 0$. So there is a $delta_{frac 15epsilon, f}$ so that $|f(x) - L| < frac 15epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$. That means $|5f(x) - 4L| < epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$.
....
And... that was exactly what we wanted!.
We will use that $delta_{frac 15epsilon, f}$ as our $delta_{epsilon, 5f}$, we get our results.
WE ARE DONE!
....
....
The difficulty is shuffling and keeping track of our variable names.
... The other difficulty is trying to write our proofs as we think of them. It really is a good idea to consider our thought process as a rough draft.
A final solution would be:
$limlimits_{xto c} f(x) = L$.
So for any $epsilon > 0$ then $frac 15epsilon > 0$ and there is a $delta$ so that whenever $|x-c| < delta$ we will have $|f(x) - L| < frac 15 epsilon$.
And thus we will have $|5(x)-5L| = 5|f(x) - L| < 5*frac 15 epsilon = epsilon$.
So $limlimits_{xto c} 5f(x) = 5L = 5limlimits_{xto c} f(x)$.
answered Jan 13 at 0:43
fleabloodfleablood
70.5k22685
$endgroup$
$begingroup$
I do understand what the definition is but I'm having trouble wrapping my head around the proofs. Since the exercises in the book don't go over arbitrary functions (f(x)).
$endgroup$
– Nick Yarn
Jan 13 at 13:16
add a comment |
$begingroup$
One thing I've found helpful in following proofs of this type is to mentally read "For any $epsilon>0$" as "For any $epsilon>0$ we choose, however small" and "a number $delta$ exists such that . . ." as "we can provide a small enough number $delta$ so that. . ."
I believe one reason people struggle so much with the $ε–δ$ definitions of limits and continuity is that the core idea behind them—of making $ε$ as small as we like by making $δ$ small enough—is absent from the actual definitions. For the purposes of the definitions it's redundant, but for the purposes of understanding what's going on it's totally essential.
answered Jan 13 at 2:58
timtfjtimtfj
2,298420
$endgroup$
add a comment |
$begingroup$
Hint:
consider taking an $epsilon$ with a specific ratio to $k$ (assuming $k neq 0$)/
That is, starting with the required $epsilon$, by the existence of the limit we can find a $delta$ s.t for $epsilon_1 := frac{epsilon}{k}$ our function is close enough to the limit. Then use that $k|a-b| = |ka-kb|$.
answered Jan 13 at 0:09
MariahMariah
1,4961618
$endgroup$
add a comment |
$begingroup$
I'm going to give you the general outline for any proof of this type. I'll quote the bits that you'd write down, and put my explanation/comments in normal text. I'm also going to assume that $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do.
For any $varepsilon > 0$,
We don't know what $delta$ needs to be yet, so we'll just leave a space there.
For any $x$ such that $|x-a|<delta$, we have $|kf(x) - kL| = |k||f(x) - L|$
Now, that's nice, because $|f(x) - L|$ is precisely what $lim_{xto c}f(x) = L$ tells us about. Specifically, it tells us that for any $barvarepsilon > 0$, there is some $bar{delta} > 0$ such that for all $x$ such that $|x-c| < bardelta$, $|f(x) - L| < barvarepsilon$.
Now, substituting that in, we see that $|kf(x) - kL| < |k|barvarepsilon$, for all $x$ such that $|x - c| < bardelta$. We need the thing on the right to be $varepsilon$, so we'll just go ahead and define (in the case $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do) $barvarepsilon := frac{varepsilon}{|k|}$ (which we can do, since our statement about $barvarepsilon$ holds for any positive $barvarepsilon$, so holds for this one in particular). Writing that properly, we'll have to go back and fill in that space we left near the start:
since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$.
Now, we can finish off the bit at the bottom, and stick some more boilerplate text on the end:
$|kf(x) - kL| = |k||f(x) - L| < |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
So, putting all of that together and sticking the standard boilerplate on the end, we've got:
For any $varepsilon > 0$, since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$. For any such $x$, we have $|kf(x) - kL| = |k||f(x) - L|< |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
And that's it. The last quoted paragraph is all you'd write down (or all you'd put in a final version, at least): everything else is just the thought process needed to produce it. Essentially all statements of this type have proofs, and thought processes, that look essentially like this.
answered Jan 13 at 0:21
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Sorry to ask but why does do we need x > 0?
$endgroup$
– Nick Yarn
Jan 13 at 0:28
$begingroup$
Oops! We don't, I edited this from a previous version, and forgot to take that out. If you're interested in seeing this approach applied to a different problem, it's here.
$endgroup$
– user3482749
Jan 13 at 0:28
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%2f3071548%2fhow-exactly-do-i-go-through-proving-a-limit-exists-with-epsilon-delta-for%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think one thing that eludes must students is that although the $delta$s and $epsilon$s are technically constants, the choice of $delta$ is dependent upon both the $epsilon$ and the function.
So we know that because God told us $limlimits_{xto c} f(x) = L$. That we know for any $epsilon > 0$ there exists a $delta_{epsilon, f}$ so the whenever $|x - c| < delta_{epsilon, f}$ then it will occur that $|f(x) -c | < epsilon$.
So why did I write this number as $delta_{epsilon, f}$ instead of just $delta$ as must texts do? I didn't have to but I did because I wanted to emphasize this particular value is determined by and "specifically tailored" for that particular $epsilon$.
So we need to prove that for any $epsilon$ there is a $delta_{epsilon, 5f}$ so that whenever $|x-c|<delta_{epsilon, 5f}$ we will have $|5f(x) - 5L| < delta_{epsilon, 5f}$. How do we find that $delta_{epsilon, 5f}$?
We $|5f(x) - 5L| = 5|f(x) - L|$ so $|5f(x) - 5L|< epsilon iff |f(x) - L| < frac 15 epsilon$. And we know that $limlimits_{xto c} f(x) = L$ so for any $epsilon > 0$ we have $delta_{epsilon f}$ that does what we want.
Well, $frac 15 epsilon > 0$. So there is a $delta_{frac 15epsilon, f}$ so that $|f(x) - L| < frac 15epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$. That means $|5f(x) - 4L| < epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$.
....
And... that was exactly what we wanted!.
We will use that $delta_{frac 15epsilon, f}$ as our $delta_{epsilon, 5f}$, we get our results.
WE ARE DONE!
....
....
The difficulty is shuffling and keeping track of our variable names.
... The other difficulty is trying to write our proofs as we think of them. It really is a good idea to consider our thought process as a rough draft.
A final solution would be:
$limlimits_{xto c} f(x) = L$.
So for any $epsilon > 0$ then $frac 15epsilon > 0$ and there is a $delta$ so that whenever $|x-c| < delta$ we will have $|f(x) - L| < frac 15 epsilon$.
And thus we will have $|5(x)-5L| = 5|f(x) - L| < 5*frac 15 epsilon = epsilon$.
So $limlimits_{xto c} 5f(x) = 5L = 5limlimits_{xto c} f(x)$.
answered Jan 13 at 0:43
fleabloodfleablood
70.5k22685
$endgroup$
$begingroup$
I do understand what the definition is but I'm having trouble wrapping my head around the proofs. Since the exercises in the book don't go over arbitrary functions (f(x)).
$endgroup$
– Nick Yarn
Jan 13 at 13:16
add a comment |
$begingroup$
I think one thing that eludes must students is that although the $delta$s and $epsilon$s are technically constants, the choice of $delta$ is dependent upon both the $epsilon$ and the function.
So we know that because God told us $limlimits_{xto c} f(x) = L$. That we know for any $epsilon > 0$ there exists a $delta_{epsilon, f}$ so the whenever $|x - c| < delta_{epsilon, f}$ then it will occur that $|f(x) -c | < epsilon$.
So why did I write this number as $delta_{epsilon, f}$ instead of just $delta$ as must texts do? I didn't have to but I did because I wanted to emphasize this particular value is determined by and "specifically tailored" for that particular $epsilon$.
So we need to prove that for any $epsilon$ there is a $delta_{epsilon, 5f}$ so that whenever $|x-c|<delta_{epsilon, 5f}$ we will have $|5f(x) - 5L| < delta_{epsilon, 5f}$. How do we find that $delta_{epsilon, 5f}$?
We $|5f(x) - 5L| = 5|f(x) - L|$ so $|5f(x) - 5L|< epsilon iff |f(x) - L| < frac 15 epsilon$. And we know that $limlimits_{xto c} f(x) = L$ so for any $epsilon > 0$ we have $delta_{epsilon f}$ that does what we want.
Well, $frac 15 epsilon > 0$. So there is a $delta_{frac 15epsilon, f}$ so that $|f(x) - L| < frac 15epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$. That means $|5f(x) - 4L| < epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$.
....
And... that was exactly what we wanted!.
We will use that $delta_{frac 15epsilon, f}$ as our $delta_{epsilon, 5f}$, we get our results.
WE ARE DONE!
....
....
The difficulty is shuffling and keeping track of our variable names.
... The other difficulty is trying to write our proofs as we think of them. It really is a good idea to consider our thought process as a rough draft.
A final solution would be:
$limlimits_{xto c} f(x) = L$.
So for any $epsilon > 0$ then $frac 15epsilon > 0$ and there is a $delta$ so that whenever $|x-c| < delta$ we will have $|f(x) - L| < frac 15 epsilon$.
And thus we will have $|5(x)-5L| = 5|f(x) - L| < 5*frac 15 epsilon = epsilon$.
So $limlimits_{xto c} 5f(x) = 5L = 5limlimits_{xto c} f(x)$.
answered Jan 13 at 0:43
fleabloodfleablood
70.5k22685
$endgroup$
$begingroup$
I do understand what the definition is but I'm having trouble wrapping my head around the proofs. Since the exercises in the book don't go over arbitrary functions (f(x)).
$endgroup$
– Nick Yarn
Jan 13 at 13:16
add a comment |
$begingroup$
I think one thing that eludes must students is that although the $delta$s and $epsilon$s are technically constants, the choice of $delta$ is dependent upon both the $epsilon$ and the function.
So we know that because God told us $limlimits_{xto c} f(x) = L$. That we know for any $epsilon > 0$ there exists a $delta_{epsilon, f}$ so the whenever $|x - c| < delta_{epsilon, f}$ then it will occur that $|f(x) -c | < epsilon$.
So why did I write this number as $delta_{epsilon, f}$ instead of just $delta$ as must texts do? I didn't have to but I did because I wanted to emphasize this particular value is determined by and "specifically tailored" for that particular $epsilon$.
So we need to prove that for any $epsilon$ there is a $delta_{epsilon, 5f}$ so that whenever $|x-c|<delta_{epsilon, 5f}$ we will have $|5f(x) - 5L| < delta_{epsilon, 5f}$. How do we find that $delta_{epsilon, 5f}$?
We $|5f(x) - 5L| = 5|f(x) - L|$ so $|5f(x) - 5L|< epsilon iff |f(x) - L| < frac 15 epsilon$. And we know that $limlimits_{xto c} f(x) = L$ so for any $epsilon > 0$ we have $delta_{epsilon f}$ that does what we want.
Well, $frac 15 epsilon > 0$. So there is a $delta_{frac 15epsilon, f}$ so that $|f(x) - L| < frac 15epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$. That means $|5f(x) - 4L| < epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$.
....
And... that was exactly what we wanted!.
We will use that $delta_{frac 15epsilon, f}$ as our $delta_{epsilon, 5f}$, we get our results.
WE ARE DONE!
....
....
The difficulty is shuffling and keeping track of our variable names.
... The other difficulty is trying to write our proofs as we think of them. It really is a good idea to consider our thought process as a rough draft.
A final solution would be:
$limlimits_{xto c} f(x) = L$.
So for any $epsilon > 0$ then $frac 15epsilon > 0$ and there is a $delta$ so that whenever $|x-c| < delta$ we will have $|f(x) - L| < frac 15 epsilon$.
And thus we will have $|5(x)-5L| = 5|f(x) - L| < 5*frac 15 epsilon = epsilon$.
So $limlimits_{xto c} 5f(x) = 5L = 5limlimits_{xto c} f(x)$.
answered Jan 13 at 0:43
fleabloodfleablood
70.5k22685
$endgroup$
I think one thing that eludes must students is that although the $delta$s and $epsilon$s are technically constants, the choice of $delta$ is dependent upon both the $epsilon$ and the function.
So we know that because God told us $limlimits_{xto c} f(x) = L$. That we know for any $epsilon > 0$ there exists a $delta_{epsilon, f}$ so the whenever $|x - c| < delta_{epsilon, f}$ then it will occur that $|f(x) -c | < epsilon$.
So why did I write this number as $delta_{epsilon, f}$ instead of just $delta$ as must texts do? I didn't have to but I did because I wanted to emphasize this particular value is determined by and "specifically tailored" for that particular $epsilon$.
So we need to prove that for any $epsilon$ there is a $delta_{epsilon, 5f}$ so that whenever $|x-c|<delta_{epsilon, 5f}$ we will have $|5f(x) - 5L| < delta_{epsilon, 5f}$. How do we find that $delta_{epsilon, 5f}$?
We $|5f(x) - 5L| = 5|f(x) - L|$ so $|5f(x) - 5L|< epsilon iff |f(x) - L| < frac 15 epsilon$. And we know that $limlimits_{xto c} f(x) = L$ so for any $epsilon > 0$ we have $delta_{epsilon f}$ that does what we want.
Well, $frac 15 epsilon > 0$. So there is a $delta_{frac 15epsilon, f}$ so that $|f(x) - L| < frac 15epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$. That means $|5f(x) - 4L| < epsilon$ when $|x-c| < delta_{frac 15epsilon, f}$.
....
And... that was exactly what we wanted!.
We will use that $delta_{frac 15epsilon, f}$ as our $delta_{epsilon, 5f}$, we get our results.
WE ARE DONE!
....
....
The difficulty is shuffling and keeping track of our variable names.
... The other difficulty is trying to write our proofs as we think of them. It really is a good idea to consider our thought process as a rough draft.
A final solution would be:
$limlimits_{xto c} f(x) = L$.
So for any $epsilon > 0$ then $frac 15epsilon > 0$ and there is a $delta$ so that whenever $|x-c| < delta$ we will have $|f(x) - L| < frac 15 epsilon$.
And thus we will have $|5(x)-5L| = 5|f(x) - L| < 5*frac 15 epsilon = epsilon$.
So $limlimits_{xto c} 5f(x) = 5L = 5limlimits_{xto c} f(x)$.
answered Jan 13 at 0:43
fleabloodfleablood
70.5k22685
answered Jan 13 at 0:43
fleabloodfleablood
70.5k22685
answered Jan 13 at 0:43
fleabloodfleablood
70.5k22685
answered Jan 13 at 0:43
fleabloodfleablood
70.5k22685
70.5k22685
$begingroup$
I do understand what the definition is but I'm having trouble wrapping my head around the proofs. Since the exercises in the book don't go over arbitrary functions (f(x)).
$endgroup$
– Nick Yarn
Jan 13 at 13:16
add a comment |
$begingroup$
I do understand what the definition is but I'm having trouble wrapping my head around the proofs. Since the exercises in the book don't go over arbitrary functions (f(x)).
$endgroup$
– Nick Yarn
Jan 13 at 13:16
$begingroup$
I do understand what the definition is but I'm having trouble wrapping my head around the proofs. Since the exercises in the book don't go over arbitrary functions (f(x)).
$endgroup$
– Nick Yarn
Jan 13 at 13:16
$begingroup$
I do understand what the definition is but I'm having trouble wrapping my head around the proofs. Since the exercises in the book don't go over arbitrary functions (f(x)).
$endgroup$
– Nick Yarn
Jan 13 at 13:16
add a comment |
$begingroup$
One thing I've found helpful in following proofs of this type is to mentally read "For any $epsilon>0$" as "For any $epsilon>0$ we choose, however small" and "a number $delta$ exists such that . . ." as "we can provide a small enough number $delta$ so that. . ."
I believe one reason people struggle so much with the $ε–δ$ definitions of limits and continuity is that the core idea behind them—of making $ε$ as small as we like by making $δ$ small enough—is absent from the actual definitions. For the purposes of the definitions it's redundant, but for the purposes of understanding what's going on it's totally essential.
answered Jan 13 at 2:58
timtfjtimtfj
2,298420
$endgroup$
add a comment |
$begingroup$
One thing I've found helpful in following proofs of this type is to mentally read "For any $epsilon>0$" as "For any $epsilon>0$ we choose, however small" and "a number $delta$ exists such that . . ." as "we can provide a small enough number $delta$ so that. . ."
I believe one reason people struggle so much with the $ε–δ$ definitions of limits and continuity is that the core idea behind them—of making $ε$ as small as we like by making $δ$ small enough—is absent from the actual definitions. For the purposes of the definitions it's redundant, but for the purposes of understanding what's going on it's totally essential.
answered Jan 13 at 2:58
timtfjtimtfj
2,298420
$endgroup$
add a comment |
$begingroup$
One thing I've found helpful in following proofs of this type is to mentally read "For any $epsilon>0$" as "For any $epsilon>0$ we choose, however small" and "a number $delta$ exists such that . . ." as "we can provide a small enough number $delta$ so that. . ."
I believe one reason people struggle so much with the $ε–δ$ definitions of limits and continuity is that the core idea behind them—of making $ε$ as small as we like by making $δ$ small enough—is absent from the actual definitions. For the purposes of the definitions it's redundant, but for the purposes of understanding what's going on it's totally essential.
answered Jan 13 at 2:58
timtfjtimtfj
2,298420
$endgroup$
One thing I've found helpful in following proofs of this type is to mentally read "For any $epsilon>0$" as "For any $epsilon>0$ we choose, however small" and "a number $delta$ exists such that . . ." as "we can provide a small enough number $delta$ so that. . ."
I believe one reason people struggle so much with the $ε–δ$ definitions of limits and continuity is that the core idea behind them—of making $ε$ as small as we like by making $δ$ small enough—is absent from the actual definitions. For the purposes of the definitions it's redundant, but for the purposes of understanding what's going on it's totally essential.
answered Jan 13 at 2:58
timtfjtimtfj
2,298420
answered Jan 13 at 2:58
timtfjtimtfj
2,298420
answered Jan 13 at 2:58
timtfjtimtfj
2,298420
answered Jan 13 at 2:58
timtfjtimtfj
2,298420
2,298420
add a comment |
add a comment |
$begingroup$
Hint:
consider taking an $epsilon$ with a specific ratio to $k$ (assuming $k neq 0$)/
That is, starting with the required $epsilon$, by the existence of the limit we can find a $delta$ s.t for $epsilon_1 := frac{epsilon}{k}$ our function is close enough to the limit. Then use that $k|a-b| = |ka-kb|$.
answered Jan 13 at 0:09
MariahMariah
1,4961618
$endgroup$
add a comment |
$begingroup$
Hint:
consider taking an $epsilon$ with a specific ratio to $k$ (assuming $k neq 0$)/
That is, starting with the required $epsilon$, by the existence of the limit we can find a $delta$ s.t for $epsilon_1 := frac{epsilon}{k}$ our function is close enough to the limit. Then use that $k|a-b| = |ka-kb|$.
answered Jan 13 at 0:09
MariahMariah
1,4961618
$endgroup$
add a comment |
$begingroup$
Hint:
consider taking an $epsilon$ with a specific ratio to $k$ (assuming $k neq 0$)/
That is, starting with the required $epsilon$, by the existence of the limit we can find a $delta$ s.t for $epsilon_1 := frac{epsilon}{k}$ our function is close enough to the limit. Then use that $k|a-b| = |ka-kb|$.
answered Jan 13 at 0:09
MariahMariah
1,4961618
$endgroup$
Hint:
consider taking an $epsilon$ with a specific ratio to $k$ (assuming $k neq 0$)/
That is, starting with the required $epsilon$, by the existence of the limit we can find a $delta$ s.t for $epsilon_1 := frac{epsilon}{k}$ our function is close enough to the limit. Then use that $k|a-b| = |ka-kb|$.
answered Jan 13 at 0:09
MariahMariah
1,4961618
answered Jan 13 at 0:09
MariahMariah
1,4961618
answered Jan 13 at 0:09
MariahMariah
1,4961618
answered Jan 13 at 0:09
MariahMariah
1,4961618
1,4961618
add a comment |
add a comment |
$begingroup$
I'm going to give you the general outline for any proof of this type. I'll quote the bits that you'd write down, and put my explanation/comments in normal text. I'm also going to assume that $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do.
For any $varepsilon > 0$,
We don't know what $delta$ needs to be yet, so we'll just leave a space there.
For any $x$ such that $|x-a|<delta$, we have $|kf(x) - kL| = |k||f(x) - L|$
Now, that's nice, because $|f(x) - L|$ is precisely what $lim_{xto c}f(x) = L$ tells us about. Specifically, it tells us that for any $barvarepsilon > 0$, there is some $bar{delta} > 0$ such that for all $x$ such that $|x-c| < bardelta$, $|f(x) - L| < barvarepsilon$.
Now, substituting that in, we see that $|kf(x) - kL| < |k|barvarepsilon$, for all $x$ such that $|x - c| < bardelta$. We need the thing on the right to be $varepsilon$, so we'll just go ahead and define (in the case $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do) $barvarepsilon := frac{varepsilon}{|k|}$ (which we can do, since our statement about $barvarepsilon$ holds for any positive $barvarepsilon$, so holds for this one in particular). Writing that properly, we'll have to go back and fill in that space we left near the start:
since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$.
Now, we can finish off the bit at the bottom, and stick some more boilerplate text on the end:
$|kf(x) - kL| = |k||f(x) - L| < |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
So, putting all of that together and sticking the standard boilerplate on the end, we've got:
For any $varepsilon > 0$, since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$. For any such $x$, we have $|kf(x) - kL| = |k||f(x) - L|< |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
And that's it. The last quoted paragraph is all you'd write down (or all you'd put in a final version, at least): everything else is just the thought process needed to produce it. Essentially all statements of this type have proofs, and thought processes, that look essentially like this.
answered Jan 13 at 0:21
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Sorry to ask but why does do we need x > 0?
$endgroup$
– Nick Yarn
Jan 13 at 0:28
$begingroup$
Oops! We don't, I edited this from a previous version, and forgot to take that out. If you're interested in seeing this approach applied to a different problem, it's here.
$endgroup$
– user3482749
Jan 13 at 0:28
add a comment |
$begingroup$
I'm going to give you the general outline for any proof of this type. I'll quote the bits that you'd write down, and put my explanation/comments in normal text. I'm also going to assume that $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do.
For any $varepsilon > 0$,
We don't know what $delta$ needs to be yet, so we'll just leave a space there.
For any $x$ such that $|x-a|<delta$, we have $|kf(x) - kL| = |k||f(x) - L|$
Now, that's nice, because $|f(x) - L|$ is precisely what $lim_{xto c}f(x) = L$ tells us about. Specifically, it tells us that for any $barvarepsilon > 0$, there is some $bar{delta} > 0$ such that for all $x$ such that $|x-c| < bardelta$, $|f(x) - L| < barvarepsilon$.
Now, substituting that in, we see that $|kf(x) - kL| < |k|barvarepsilon$, for all $x$ such that $|x - c| < bardelta$. We need the thing on the right to be $varepsilon$, so we'll just go ahead and define (in the case $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do) $barvarepsilon := frac{varepsilon}{|k|}$ (which we can do, since our statement about $barvarepsilon$ holds for any positive $barvarepsilon$, so holds for this one in particular). Writing that properly, we'll have to go back and fill in that space we left near the start:
since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$.
Now, we can finish off the bit at the bottom, and stick some more boilerplate text on the end:
$|kf(x) - kL| = |k||f(x) - L| < |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
So, putting all of that together and sticking the standard boilerplate on the end, we've got:
For any $varepsilon > 0$, since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$. For any such $x$, we have $|kf(x) - kL| = |k||f(x) - L|< |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
And that's it. The last quoted paragraph is all you'd write down (or all you'd put in a final version, at least): everything else is just the thought process needed to produce it. Essentially all statements of this type have proofs, and thought processes, that look essentially like this.
answered Jan 13 at 0:21
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Sorry to ask but why does do we need x > 0?
$endgroup$
– Nick Yarn
Jan 13 at 0:28
$begingroup$
Oops! We don't, I edited this from a previous version, and forgot to take that out. If you're interested in seeing this approach applied to a different problem, it's here.
$endgroup$
– user3482749
Jan 13 at 0:28
add a comment |
$begingroup$
I'm going to give you the general outline for any proof of this type. I'll quote the bits that you'd write down, and put my explanation/comments in normal text. I'm also going to assume that $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do.
For any $varepsilon > 0$,
We don't know what $delta$ needs to be yet, so we'll just leave a space there.
For any $x$ such that $|x-a|<delta$, we have $|kf(x) - kL| = |k||f(x) - L|$
Now, that's nice, because $|f(x) - L|$ is precisely what $lim_{xto c}f(x) = L$ tells us about. Specifically, it tells us that for any $barvarepsilon > 0$, there is some $bar{delta} > 0$ such that for all $x$ such that $|x-c| < bardelta$, $|f(x) - L| < barvarepsilon$.
Now, substituting that in, we see that $|kf(x) - kL| < |k|barvarepsilon$, for all $x$ such that $|x - c| < bardelta$. We need the thing on the right to be $varepsilon$, so we'll just go ahead and define (in the case $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do) $barvarepsilon := frac{varepsilon}{|k|}$ (which we can do, since our statement about $barvarepsilon$ holds for any positive $barvarepsilon$, so holds for this one in particular). Writing that properly, we'll have to go back and fill in that space we left near the start:
since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$.
Now, we can finish off the bit at the bottom, and stick some more boilerplate text on the end:
$|kf(x) - kL| = |k||f(x) - L| < |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
So, putting all of that together and sticking the standard boilerplate on the end, we've got:
For any $varepsilon > 0$, since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$. For any such $x$, we have $|kf(x) - kL| = |k||f(x) - L|< |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
And that's it. The last quoted paragraph is all you'd write down (or all you'd put in a final version, at least): everything else is just the thought process needed to produce it. Essentially all statements of this type have proofs, and thought processes, that look essentially like this.
answered Jan 13 at 0:21
user3482749user3482749
4,266919
$endgroup$
I'm going to give you the general outline for any proof of this type. I'll quote the bits that you'd write down, and put my explanation/comments in normal text. I'm also going to assume that $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do.
For any $varepsilon > 0$,
We don't know what $delta$ needs to be yet, so we'll just leave a space there.
For any $x$ such that $|x-a|<delta$, we have $|kf(x) - kL| = |k||f(x) - L|$
Now, that's nice, because $|f(x) - L|$ is precisely what $lim_{xto c}f(x) = L$ tells us about. Specifically, it tells us that for any $barvarepsilon > 0$, there is some $bar{delta} > 0$ such that for all $x$ such that $|x-c| < bardelta$, $|f(x) - L| < barvarepsilon$.
Now, substituting that in, we see that $|kf(x) - kL| < |k|barvarepsilon$, for all $x$ such that $|x - c| < bardelta$. We need the thing on the right to be $varepsilon$, so we'll just go ahead and define (in the case $k neq 0$: the $k = 0$ case is easy, and I'll leave it for you to do) $barvarepsilon := frac{varepsilon}{|k|}$ (which we can do, since our statement about $barvarepsilon$ holds for any positive $barvarepsilon$, so holds for this one in particular). Writing that properly, we'll have to go back and fill in that space we left near the start:
since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$.
Now, we can finish off the bit at the bottom, and stick some more boilerplate text on the end:
$|kf(x) - kL| = |k||f(x) - L| < |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
So, putting all of that together and sticking the standard boilerplate on the end, we've got:
For any $varepsilon > 0$, since $lim_{xto c}f(x) = L$, and since $frac{varepsilon}{|k|} > 0$, there is some $delta > 0$ such that for all $x$ such that $|x - c| <delta$, $|f(x) - L| < frac{varepsilon}{|k|}$. For any such $x$, we have $|kf(x) - kL| = |k||f(x) - L|< |k|frac{varepsilon}{|k|} = varepsilon$. Thus, $lim_{xto c}kf(x) = kL$, as required.
And that's it. The last quoted paragraph is all you'd write down (or all you'd put in a final version, at least): everything else is just the thought process needed to produce it. Essentially all statements of this type have proofs, and thought processes, that look essentially like this.
answered Jan 13 at 0:21
user3482749user3482749
4,266919
edited Jan 13 at 0:29
answered Jan 13 at 0:21
user3482749user3482749
4,266919
answered Jan 13 at 0:21
user3482749user3482749
4,266919
answered Jan 13 at 0:21
user3482749user3482749
4,266919
4,266919
$begingroup$
Sorry to ask but why does do we need x > 0?
$endgroup$
– Nick Yarn
Jan 13 at 0:28
$begingroup$
Oops! We don't, I edited this from a previous version, and forgot to take that out. If you're interested in seeing this approach applied to a different problem, it's here.
$endgroup$
– user3482749
Jan 13 at 0:28
add a comment |
$begingroup$
Sorry to ask but why does do we need x > 0?
$endgroup$
– Nick Yarn
Jan 13 at 0:28
$begingroup$
Oops! We don't, I edited this from a previous version, and forgot to take that out. If you're interested in seeing this approach applied to a different problem, it's here.
$endgroup$
– user3482749
Jan 13 at 0:28
$begingroup$
Sorry to ask but why does do we need x > 0?
$endgroup$
– Nick Yarn
Jan 13 at 0:28
$begingroup$
Sorry to ask but why does do we need x > 0?
$endgroup$
– Nick Yarn
Jan 13 at 0:28
$begingroup$
Oops! We don't, I edited this from a previous version, and forgot to take that out. If you're interested in seeing this approach applied to a different problem, it's here.
$endgroup$
– user3482749
Jan 13 at 0:28
$begingroup$
Oops! We don't, I edited this from a previous version, and forgot to take that out. If you're interested in seeing this approach applied to a different problem, it's here.
$endgroup$
– user3482749
Jan 13 at 0:28
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%2f3071548%2fhow-exactly-do-i-go-through-proving-a-limit-exists-with-epsilon-delta-for%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
