Mixing
Calculating the Limit of a Function without use of L'Hospital's Rule or Power Series
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I came across this question the other day and have been trying to solve it by using some simple algebraic manipulation without really delving into L'Hospital's Rule or the Power Series as I have just started learning limit calculations.
We needed to find :
$$lim_{x to 0} frac {xcos x - sin x}{x^2sin x}$$
I approached this problem in two different ways and know what the flaw is, however I have been unable to justify why this is so.
Let $$f(x) = frac {xcos x - sin x}{x^2sin x}$$
Therefore, dividing by $x$,
$$f(x) = frac {cos x - frac{sin x}{x}}{xsin x}$$
Using standard limit properties,
$$lim_{x to 0}f(x) = frac{lim_{x to 0}cos x - lim_{x to 0}frac{sin x}{x}}{lim_{x to 0}xsin x}$$
Since $$lim_{x to 0} frac {sin x}{x}=1$$
$$lim_{x to 0}f(x)= frac{lim_{x to 0}cos x-1}{lim_{x to 0}xsin x}$$
Rewriting the above as $$lim_{x to 0}frac{(cos x -1)x}{x^2sin x}$$ and using the fact that $lim_{x to 0} frac {sin x}{x}=1$ and $lim_{x to 0}frac{cos x -1}{x^2}= -frac{1}{2}$,
we get $$lim_{x to 0}f(x)=-frac{1}{2}$$
I know that the answer is wrong although I am not able to understand why.
I believe it is because I cannot combine the numerator and denominator into a single limit function. Using a similar trick, I also obtained the limit to be $-frac{3}{8}$.
Questions:
1) Could someone please explain why combining the numerator and denominator into a single limit is wrong? (The reason I even went ahead with such a manipulation was, we are allowed to separate the numerator and denominator while expanding the limit of a rational function so I felt that the reverse should also work).
2) As you can notice, I have not used L'Hospital's Rules or Power Series expansion of $sin x $and $cos x$. When I used L'Hospital's Rule, I noticed that I needed to go upto the third or fourth derivative to get rid of the $frac{0}{0}$ indeterminate form. So would there be a better way of approaching such limits?
Thank You.
limits power-series limits-without-lhopital
asked Jan 28 at 16:36
Aditya SriramAditya Sriram
132
$endgroup$
add a comment |
$begingroup$
I came across this question the other day and have been trying to solve it by using some simple algebraic manipulation without really delving into L'Hospital's Rule or the Power Series as I have just started learning limit calculations.
We needed to find :
$$lim_{x to 0} frac {xcos x - sin x}{x^2sin x}$$
I approached this problem in two different ways and know what the flaw is, however I have been unable to justify why this is so.
Let $$f(x) = frac {xcos x - sin x}{x^2sin x}$$
Therefore, dividing by $x$,
$$f(x) = frac {cos x - frac{sin x}{x}}{xsin x}$$
Using standard limit properties,
$$lim_{x to 0}f(x) = frac{lim_{x to 0}cos x - lim_{x to 0}frac{sin x}{x}}{lim_{x to 0}xsin x}$$
Since $$lim_{x to 0} frac {sin x}{x}=1$$
$$lim_{x to 0}f(x)= frac{lim_{x to 0}cos x-1}{lim_{x to 0}xsin x}$$
Rewriting the above as $$lim_{x to 0}frac{(cos x -1)x}{x^2sin x}$$ and using the fact that $lim_{x to 0} frac {sin x}{x}=1$ and $lim_{x to 0}frac{cos x -1}{x^2}= -frac{1}{2}$,
we get $$lim_{x to 0}f(x)=-frac{1}{2}$$
I know that the answer is wrong although I am not able to understand why.
I believe it is because I cannot combine the numerator and denominator into a single limit function. Using a similar trick, I also obtained the limit to be $-frac{3}{8}$.
Questions:
1) Could someone please explain why combining the numerator and denominator into a single limit is wrong? (The reason I even went ahead with such a manipulation was, we are allowed to separate the numerator and denominator while expanding the limit of a rational function so I felt that the reverse should also work).
2) As you can notice, I have not used L'Hospital's Rules or Power Series expansion of $sin x $and $cos x$. When I used L'Hospital's Rule, I noticed that I needed to go upto the third or fourth derivative to get rid of the $frac{0}{0}$ indeterminate form. So would there be a better way of approaching such limits?
Thank You.
limits power-series limits-without-lhopital
asked Jan 28 at 16:36
Aditya SriramAditya Sriram
132
$endgroup$
$begingroup$
$limfrac fg=frac{lim f}{lim g}$ iff $lim f,lim g$ exist and $lim gne0$, so your second step is wrong. Say it wasn't, then your next step falters when you selectively apply the limit to $dfrac{sin x}x$ in the numerator but not to $cos x$.
$endgroup$
– Shubham Johri
Jan 28 at 16:59
$begingroup$
@ShubhamJohri: there is no faltering in second step. Convince yourself that the following is true. If $g(x) to L$ as $xto a$ then $$lim_{xto a} {f(x) pm g(x) } =lim_{xto a} f(x) pm L$$ See more details at math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Jan 29 at 2:32
add a comment |
$begingroup$
I came across this question the other day and have been trying to solve it by using some simple algebraic manipulation without really delving into L'Hospital's Rule or the Power Series as I have just started learning limit calculations.
We needed to find :
$$lim_{x to 0} frac {xcos x - sin x}{x^2sin x}$$
I approached this problem in two different ways and know what the flaw is, however I have been unable to justify why this is so.
Let $$f(x) = frac {xcos x - sin x}{x^2sin x}$$
Therefore, dividing by $x$,
$$f(x) = frac {cos x - frac{sin x}{x}}{xsin x}$$
Using standard limit properties,
$$lim_{x to 0}f(x) = frac{lim_{x to 0}cos x - lim_{x to 0}frac{sin x}{x}}{lim_{x to 0}xsin x}$$
Since $$lim_{x to 0} frac {sin x}{x}=1$$
$$lim_{x to 0}f(x)= frac{lim_{x to 0}cos x-1}{lim_{x to 0}xsin x}$$
Rewriting the above as $$lim_{x to 0}frac{(cos x -1)x}{x^2sin x}$$ and using the fact that $lim_{x to 0} frac {sin x}{x}=1$ and $lim_{x to 0}frac{cos x -1}{x^2}= -frac{1}{2}$,
we get $$lim_{x to 0}f(x)=-frac{1}{2}$$
I know that the answer is wrong although I am not able to understand why.
I believe it is because I cannot combine the numerator and denominator into a single limit function. Using a similar trick, I also obtained the limit to be $-frac{3}{8}$.
Questions:
1) Could someone please explain why combining the numerator and denominator into a single limit is wrong? (The reason I even went ahead with such a manipulation was, we are allowed to separate the numerator and denominator while expanding the limit of a rational function so I felt that the reverse should also work).
2) As you can notice, I have not used L'Hospital's Rules or Power Series expansion of $sin x $and $cos x$. When I used L'Hospital's Rule, I noticed that I needed to go upto the third or fourth derivative to get rid of the $frac{0}{0}$ indeterminate form. So would there be a better way of approaching such limits?
Thank You.
limits power-series limits-without-lhopital
asked Jan 28 at 16:36
Aditya SriramAditya Sriram
132
$endgroup$
I came across this question the other day and have been trying to solve it by using some simple algebraic manipulation without really delving into L'Hospital's Rule or the Power Series as I have just started learning limit calculations.
We needed to find :
$$lim_{x to 0} frac {xcos x - sin x}{x^2sin x}$$
I approached this problem in two different ways and know what the flaw is, however I have been unable to justify why this is so.
Let $$f(x) = frac {xcos x - sin x}{x^2sin x}$$
Therefore, dividing by $x$,
$$f(x) = frac {cos x - frac{sin x}{x}}{xsin x}$$
Using standard limit properties,
$$lim_{x to 0}f(x) = frac{lim_{x to 0}cos x - lim_{x to 0}frac{sin x}{x}}{lim_{x to 0}xsin x}$$
Since $$lim_{x to 0} frac {sin x}{x}=1$$
$$lim_{x to 0}f(x)= frac{lim_{x to 0}cos x-1}{lim_{x to 0}xsin x}$$
Rewriting the above as $$lim_{x to 0}frac{(cos x -1)x}{x^2sin x}$$ and using the fact that $lim_{x to 0} frac {sin x}{x}=1$ and $lim_{x to 0}frac{cos x -1}{x^2}= -frac{1}{2}$,
we get $$lim_{x to 0}f(x)=-frac{1}{2}$$
I know that the answer is wrong although I am not able to understand why.
I believe it is because I cannot combine the numerator and denominator into a single limit function. Using a similar trick, I also obtained the limit to be $-frac{3}{8}$.
Questions:
1) Could someone please explain why combining the numerator and denominator into a single limit is wrong? (The reason I even went ahead with such a manipulation was, we are allowed to separate the numerator and denominator while expanding the limit of a rational function so I felt that the reverse should also work).
2) As you can notice, I have not used L'Hospital's Rules or Power Series expansion of $sin x $and $cos x$. When I used L'Hospital's Rule, I noticed that I needed to go upto the third or fourth derivative to get rid of the $frac{0}{0}$ indeterminate form. So would there be a better way of approaching such limits?
Thank You.
limits power-series limits-without-lhopital
limits power-series limits-without-lhopital
asked Jan 28 at 16:36
Aditya SriramAditya Sriram
132
asked Jan 28 at 16:36
Aditya SriramAditya Sriram
132
asked Jan 28 at 16:36
Aditya SriramAditya Sriram
132
asked Jan 28 at 16:36
Aditya SriramAditya Sriram
132
asked Jan 28 at 16:36
Aditya SriramAditya Sriram
132
132
$begingroup$
$limfrac fg=frac{lim f}{lim g}$ iff $lim f,lim g$ exist and $lim gne0$, so your second step is wrong. Say it wasn't, then your next step falters when you selectively apply the limit to $dfrac{sin x}x$ in the numerator but not to $cos x$.
$endgroup$
– Shubham Johri
Jan 28 at 16:59
$begingroup$
@ShubhamJohri: there is no faltering in second step. Convince yourself that the following is true. If $g(x) to L$ as $xto a$ then $$lim_{xto a} {f(x) pm g(x) } =lim_{xto a} f(x) pm L$$ See more details at math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Jan 29 at 2:32
add a comment |
$begingroup$
$limfrac fg=frac{lim f}{lim g}$ iff $lim f,lim g$ exist and $lim gne0$, so your second step is wrong. Say it wasn't, then your next step falters when you selectively apply the limit to $dfrac{sin x}x$ in the numerator but not to $cos x$.
$endgroup$
– Shubham Johri
Jan 28 at 16:59
$begingroup$
@ShubhamJohri: there is no faltering in second step. Convince yourself that the following is true. If $g(x) to L$ as $xto a$ then $$lim_{xto a} {f(x) pm g(x) } =lim_{xto a} f(x) pm L$$ See more details at math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Jan 29 at 2:32
$begingroup$
$limfrac fg=frac{lim f}{lim g}$ iff $lim f,lim g$ exist and $lim gne0$, so your second step is wrong. Say it wasn't, then your next step falters when you selectively apply the limit to $dfrac{sin x}x$ in the numerator but not to $cos x$.
$endgroup$
– Shubham Johri
Jan 28 at 16:59
$begingroup$
$limfrac fg=frac{lim f}{lim g}$ iff $lim f,lim g$ exist and $lim gne0$, so your second step is wrong. Say it wasn't, then your next step falters when you selectively apply the limit to $dfrac{sin x}x$ in the numerator but not to $cos x$.
$endgroup$
– Shubham Johri
Jan 28 at 16:59
$begingroup$
@ShubhamJohri: there is no faltering in second step. Convince yourself that the following is true. If $g(x) to L$ as $xto a$ then $$lim_{xto a} {f(x) pm g(x) } =lim_{xto a} f(x) pm L$$ See more details at math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Jan 29 at 2:32
$begingroup$
@ShubhamJohri: there is no faltering in second step. Convince yourself that the following is true. If $g(x) to L$ as $xto a$ then $$lim_{xto a} {f(x) pm g(x) } =lim_{xto a} f(x) pm L$$ See more details at math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Jan 29 at 2:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First, note that both of your answers are incorrect: the actual answer is $frac{-1}{3}$.
For your questions:
- The problem is actually earlier than that: it is not necessarily the case that $lim frac{f(x)}{g(x)} = frac{lim f(x)}{lim g(x)}$ unless both limits exist, and $lim g(x) neq 0$. In this case, that latter condition isn't satisfied, so this method does not work. In particular, when you write:
$$lim_{xto 0}f(x) = dfrac{lim_{xto 0}cos(x) - lim_{xto 0}frac{sin(x)}{x}}{lim_{xto 0} xsin(x)},$$
what you've actually written is $limlimits_{xto 0}f(x) = frac{mbox{[something]}}{0}$, which is obviously nonsense, and so everything after that is necessarily flawed.
- L'Hôpital's rule (he's not a hospital!) is how I'd approach this. There might be a trick to do it quickly, but it's not obvious if there is.
answered Jan 28 at 16:58
user3482749user3482749
4,3311119
$endgroup$
$begingroup$
Thank you for pointing out my mistake.
$endgroup$
– Aditya Sriram
Jan 28 at 17:03
add a comment |
$begingroup$
Your approach does not use the laws of limits in proper manner. Each of those laws which deal with limit of sum/difference/product/quotient of two functions has certain restrictions which must be checked before applying them.
In the current question the issue is with application of the quotient rule. You should note that the rule applies only when the limit of denominator is non-zero. But here limit of denominator $xsin x$ is $0$ and thus your first step is wrong.
Further note that the question does require the use of L'Hospital's Rule or Taylor series. Using the limit $limlimits _{xto 0}dfrac{sin x} {x} =1$ you can replace the denominator $x^2sin x$ (as given in original problem) by $x^3$ and the numerator should be split as $$xcos x-x+(x-sin x) $$ so that the fraction under limit can be replaced by $$frac{x-sin x} {x^3}-frac{1-cos x} {x^2}$$ The second fraction tends to $1/2$ (via algebraic manipulation) and the first one tends to $1/6$ (via L'Hospital's Rule or Taylor series) and thus the desired limit is $1/6-1/2=-1/3$.
answered Jan 29 at 2:25
Paramanand SinghParamanand Singh
51.1k557168
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add a comment |
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1) If I understand your question, you are conjecturing that:
$$
lim_{x to a} frac{f(x)}{g(x)}= frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}
$$
This is wrong. You may try to prove it, but as a simple counterexample should suffice:
Let $f(x)=x$ and $g(x) =x$. Then:
$$
lim_{x to a} frac{f(x)}{g(x)}= lim_{x to a} frac{x}{x} = 1
$$
And the result holds even for $a=0$.
but:
$$
frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}= frac{a}{a}
$$
And the last one is undetermined for $a =0$, since $a/a$ is not within a limit expression anymore. This would actually hold if ${lim_{x to a} g(x)}neq 0$, but this exception was precisely the case you had in hand.
2) You have a rather complicate expression there. I don't see any elegant/short solution to it, but using power series seems pretty much fair and straight forward.
There could be some property related to limits and some known transform such as Initial value theorem from Laplace's transform, but that wouldn't enter into my book as a "better" solution.
answered Jan 28 at 16:52
MefiticoMefitico
1,182218
$endgroup$
$begingroup$
The rule you mention at the beginning is actually correct but has a hypothesis that limit of $g(x) $ is non-zero.
$endgroup$
– Paramanand Singh
Jan 29 at 2:27
$begingroup$
@ParamanandSingh : Yes, but that's precisely the case in which OP tried to use the rule.
$endgroup$
– Mefitico
Jan 29 at 2:45
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, note that both of your answers are incorrect: the actual answer is $frac{-1}{3}$.
For your questions:
- The problem is actually earlier than that: it is not necessarily the case that $lim frac{f(x)}{g(x)} = frac{lim f(x)}{lim g(x)}$ unless both limits exist, and $lim g(x) neq 0$. In this case, that latter condition isn't satisfied, so this method does not work. In particular, when you write:
$$lim_{xto 0}f(x) = dfrac{lim_{xto 0}cos(x) - lim_{xto 0}frac{sin(x)}{x}}{lim_{xto 0} xsin(x)},$$
what you've actually written is $limlimits_{xto 0}f(x) = frac{mbox{[something]}}{0}$, which is obviously nonsense, and so everything after that is necessarily flawed.
- L'Hôpital's rule (he's not a hospital!) is how I'd approach this. There might be a trick to do it quickly, but it's not obvious if there is.
answered Jan 28 at 16:58
user3482749user3482749
4,3311119
$endgroup$
$begingroup$
Thank you for pointing out my mistake.
$endgroup$
– Aditya Sriram
Jan 28 at 17:03
add a comment |
$begingroup$
First, note that both of your answers are incorrect: the actual answer is $frac{-1}{3}$.
For your questions:
- The problem is actually earlier than that: it is not necessarily the case that $lim frac{f(x)}{g(x)} = frac{lim f(x)}{lim g(x)}$ unless both limits exist, and $lim g(x) neq 0$. In this case, that latter condition isn't satisfied, so this method does not work. In particular, when you write:
$$lim_{xto 0}f(x) = dfrac{lim_{xto 0}cos(x) - lim_{xto 0}frac{sin(x)}{x}}{lim_{xto 0} xsin(x)},$$
what you've actually written is $limlimits_{xto 0}f(x) = frac{mbox{[something]}}{0}$, which is obviously nonsense, and so everything after that is necessarily flawed.
- L'Hôpital's rule (he's not a hospital!) is how I'd approach this. There might be a trick to do it quickly, but it's not obvious if there is.
answered Jan 28 at 16:58
user3482749user3482749
4,3311119
$endgroup$
$begingroup$
Thank you for pointing out my mistake.
$endgroup$
– Aditya Sriram
Jan 28 at 17:03
add a comment |
$begingroup$
First, note that both of your answers are incorrect: the actual answer is $frac{-1}{3}$.
For your questions:
- The problem is actually earlier than that: it is not necessarily the case that $lim frac{f(x)}{g(x)} = frac{lim f(x)}{lim g(x)}$ unless both limits exist, and $lim g(x) neq 0$. In this case, that latter condition isn't satisfied, so this method does not work. In particular, when you write:
$$lim_{xto 0}f(x) = dfrac{lim_{xto 0}cos(x) - lim_{xto 0}frac{sin(x)}{x}}{lim_{xto 0} xsin(x)},$$
what you've actually written is $limlimits_{xto 0}f(x) = frac{mbox{[something]}}{0}$, which is obviously nonsense, and so everything after that is necessarily flawed.
- L'Hôpital's rule (he's not a hospital!) is how I'd approach this. There might be a trick to do it quickly, but it's not obvious if there is.
answered Jan 28 at 16:58
user3482749user3482749
4,3311119
$endgroup$
First, note that both of your answers are incorrect: the actual answer is $frac{-1}{3}$.
For your questions:
- The problem is actually earlier than that: it is not necessarily the case that $lim frac{f(x)}{g(x)} = frac{lim f(x)}{lim g(x)}$ unless both limits exist, and $lim g(x) neq 0$. In this case, that latter condition isn't satisfied, so this method does not work. In particular, when you write:
$$lim_{xto 0}f(x) = dfrac{lim_{xto 0}cos(x) - lim_{xto 0}frac{sin(x)}{x}}{lim_{xto 0} xsin(x)},$$
what you've actually written is $limlimits_{xto 0}f(x) = frac{mbox{[something]}}{0}$, which is obviously nonsense, and so everything after that is necessarily flawed.
- L'Hôpital's rule (he's not a hospital!) is how I'd approach this. There might be a trick to do it quickly, but it's not obvious if there is.
answered Jan 28 at 16:58
user3482749user3482749
4,3311119
answered Jan 28 at 16:58
user3482749user3482749
4,3311119
answered Jan 28 at 16:58
user3482749user3482749
4,3311119
answered Jan 28 at 16:58
user3482749user3482749
4,3311119
4,3311119
$begingroup$
Thank you for pointing out my mistake.
$endgroup$
– Aditya Sriram
Jan 28 at 17:03
add a comment |
$begingroup$
Thank you for pointing out my mistake.
$endgroup$
– Aditya Sriram
Jan 28 at 17:03
$begingroup$
Thank you for pointing out my mistake.
$endgroup$
– Aditya Sriram
Jan 28 at 17:03
$begingroup$
Thank you for pointing out my mistake.
$endgroup$
– Aditya Sriram
Jan 28 at 17:03
add a comment |
$begingroup$
Your approach does not use the laws of limits in proper manner. Each of those laws which deal with limit of sum/difference/product/quotient of two functions has certain restrictions which must be checked before applying them.
In the current question the issue is with application of the quotient rule. You should note that the rule applies only when the limit of denominator is non-zero. But here limit of denominator $xsin x$ is $0$ and thus your first step is wrong.
Further note that the question does require the use of L'Hospital's Rule or Taylor series. Using the limit $limlimits _{xto 0}dfrac{sin x} {x} =1$ you can replace the denominator $x^2sin x$ (as given in original problem) by $x^3$ and the numerator should be split as $$xcos x-x+(x-sin x) $$ so that the fraction under limit can be replaced by $$frac{x-sin x} {x^3}-frac{1-cos x} {x^2}$$ The second fraction tends to $1/2$ (via algebraic manipulation) and the first one tends to $1/6$ (via L'Hospital's Rule or Taylor series) and thus the desired limit is $1/6-1/2=-1/3$.
answered Jan 29 at 2:25
Paramanand SinghParamanand Singh
51.1k557168
$endgroup$
add a comment |
$begingroup$
Your approach does not use the laws of limits in proper manner. Each of those laws which deal with limit of sum/difference/product/quotient of two functions has certain restrictions which must be checked before applying them.
In the current question the issue is with application of the quotient rule. You should note that the rule applies only when the limit of denominator is non-zero. But here limit of denominator $xsin x$ is $0$ and thus your first step is wrong.
Further note that the question does require the use of L'Hospital's Rule or Taylor series. Using the limit $limlimits _{xto 0}dfrac{sin x} {x} =1$ you can replace the denominator $x^2sin x$ (as given in original problem) by $x^3$ and the numerator should be split as $$xcos x-x+(x-sin x) $$ so that the fraction under limit can be replaced by $$frac{x-sin x} {x^3}-frac{1-cos x} {x^2}$$ The second fraction tends to $1/2$ (via algebraic manipulation) and the first one tends to $1/6$ (via L'Hospital's Rule or Taylor series) and thus the desired limit is $1/6-1/2=-1/3$.
answered Jan 29 at 2:25
Paramanand SinghParamanand Singh
51.1k557168
$endgroup$
add a comment |
$begingroup$
Your approach does not use the laws of limits in proper manner. Each of those laws which deal with limit of sum/difference/product/quotient of two functions has certain restrictions which must be checked before applying them.
In the current question the issue is with application of the quotient rule. You should note that the rule applies only when the limit of denominator is non-zero. But here limit of denominator $xsin x$ is $0$ and thus your first step is wrong.
Further note that the question does require the use of L'Hospital's Rule or Taylor series. Using the limit $limlimits _{xto 0}dfrac{sin x} {x} =1$ you can replace the denominator $x^2sin x$ (as given in original problem) by $x^3$ and the numerator should be split as $$xcos x-x+(x-sin x) $$ so that the fraction under limit can be replaced by $$frac{x-sin x} {x^3}-frac{1-cos x} {x^2}$$ The second fraction tends to $1/2$ (via algebraic manipulation) and the first one tends to $1/6$ (via L'Hospital's Rule or Taylor series) and thus the desired limit is $1/6-1/2=-1/3$.
answered Jan 29 at 2:25
Paramanand SinghParamanand Singh
51.1k557168
$endgroup$
Your approach does not use the laws of limits in proper manner. Each of those laws which deal with limit of sum/difference/product/quotient of two functions has certain restrictions which must be checked before applying them.
In the current question the issue is with application of the quotient rule. You should note that the rule applies only when the limit of denominator is non-zero. But here limit of denominator $xsin x$ is $0$ and thus your first step is wrong.
Further note that the question does require the use of L'Hospital's Rule or Taylor series. Using the limit $limlimits _{xto 0}dfrac{sin x} {x} =1$ you can replace the denominator $x^2sin x$ (as given in original problem) by $x^3$ and the numerator should be split as $$xcos x-x+(x-sin x) $$ so that the fraction under limit can be replaced by $$frac{x-sin x} {x^3}-frac{1-cos x} {x^2}$$ The second fraction tends to $1/2$ (via algebraic manipulation) and the first one tends to $1/6$ (via L'Hospital's Rule or Taylor series) and thus the desired limit is $1/6-1/2=-1/3$.
answered Jan 29 at 2:25
Paramanand SinghParamanand Singh
51.1k557168
answered Jan 29 at 2:25
Paramanand SinghParamanand Singh
51.1k557168
answered Jan 29 at 2:25
Paramanand SinghParamanand Singh
51.1k557168
answered Jan 29 at 2:25
Paramanand SinghParamanand Singh
51.1k557168
51.1k557168
add a comment |
add a comment |
$begingroup$
1) If I understand your question, you are conjecturing that:
$$
lim_{x to a} frac{f(x)}{g(x)}= frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}
$$
This is wrong. You may try to prove it, but as a simple counterexample should suffice:
Let $f(x)=x$ and $g(x) =x$. Then:
$$
lim_{x to a} frac{f(x)}{g(x)}= lim_{x to a} frac{x}{x} = 1
$$
And the result holds even for $a=0$.
but:
$$
frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}= frac{a}{a}
$$
And the last one is undetermined for $a =0$, since $a/a$ is not within a limit expression anymore. This would actually hold if ${lim_{x to a} g(x)}neq 0$, but this exception was precisely the case you had in hand.
2) You have a rather complicate expression there. I don't see any elegant/short solution to it, but using power series seems pretty much fair and straight forward.
There could be some property related to limits and some known transform such as Initial value theorem from Laplace's transform, but that wouldn't enter into my book as a "better" solution.
answered Jan 28 at 16:52
MefiticoMefitico
1,182218
$endgroup$
$begingroup$
The rule you mention at the beginning is actually correct but has a hypothesis that limit of $g(x) $ is non-zero.
$endgroup$
– Paramanand Singh
Jan 29 at 2:27
$begingroup$
@ParamanandSingh : Yes, but that's precisely the case in which OP tried to use the rule.
$endgroup$
– Mefitico
Jan 29 at 2:45
add a comment |
$begingroup$
1) If I understand your question, you are conjecturing that:
$$
lim_{x to a} frac{f(x)}{g(x)}= frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}
$$
This is wrong. You may try to prove it, but as a simple counterexample should suffice:
Let $f(x)=x$ and $g(x) =x$. Then:
$$
lim_{x to a} frac{f(x)}{g(x)}= lim_{x to a} frac{x}{x} = 1
$$
And the result holds even for $a=0$.
but:
$$
frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}= frac{a}{a}
$$
And the last one is undetermined for $a =0$, since $a/a$ is not within a limit expression anymore. This would actually hold if ${lim_{x to a} g(x)}neq 0$, but this exception was precisely the case you had in hand.
2) You have a rather complicate expression there. I don't see any elegant/short solution to it, but using power series seems pretty much fair and straight forward.
There could be some property related to limits and some known transform such as Initial value theorem from Laplace's transform, but that wouldn't enter into my book as a "better" solution.
answered Jan 28 at 16:52
MefiticoMefitico
1,182218
$endgroup$
$begingroup$
The rule you mention at the beginning is actually correct but has a hypothesis that limit of $g(x) $ is non-zero.
$endgroup$
– Paramanand Singh
Jan 29 at 2:27
$begingroup$
@ParamanandSingh : Yes, but that's precisely the case in which OP tried to use the rule.
$endgroup$
– Mefitico
Jan 29 at 2:45
add a comment |
$begingroup$
1) If I understand your question, you are conjecturing that:
$$
lim_{x to a} frac{f(x)}{g(x)}= frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}
$$
This is wrong. You may try to prove it, but as a simple counterexample should suffice:
Let $f(x)=x$ and $g(x) =x$. Then:
$$
lim_{x to a} frac{f(x)}{g(x)}= lim_{x to a} frac{x}{x} = 1
$$
And the result holds even for $a=0$.
but:
$$
frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}= frac{a}{a}
$$
And the last one is undetermined for $a =0$, since $a/a$ is not within a limit expression anymore. This would actually hold if ${lim_{x to a} g(x)}neq 0$, but this exception was precisely the case you had in hand.
2) You have a rather complicate expression there. I don't see any elegant/short solution to it, but using power series seems pretty much fair and straight forward.
There could be some property related to limits and some known transform such as Initial value theorem from Laplace's transform, but that wouldn't enter into my book as a "better" solution.
answered Jan 28 at 16:52
MefiticoMefitico
1,182218
$endgroup$
1) If I understand your question, you are conjecturing that:
$$
lim_{x to a} frac{f(x)}{g(x)}= frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}
$$
This is wrong. You may try to prove it, but as a simple counterexample should suffice:
Let $f(x)=x$ and $g(x) =x$. Then:
$$
lim_{x to a} frac{f(x)}{g(x)}= lim_{x to a} frac{x}{x} = 1
$$
And the result holds even for $a=0$.
but:
$$
frac{lim_{x to a} f(x)}{lim_{x to a} g(x)}= frac{a}{a}
$$
And the last one is undetermined for $a =0$, since $a/a$ is not within a limit expression anymore. This would actually hold if ${lim_{x to a} g(x)}neq 0$, but this exception was precisely the case you had in hand.
2) You have a rather complicate expression there. I don't see any elegant/short solution to it, but using power series seems pretty much fair and straight forward.
There could be some property related to limits and some known transform such as Initial value theorem from Laplace's transform, but that wouldn't enter into my book as a "better" solution.
answered Jan 28 at 16:52
MefiticoMefitico
1,182218
edited Jan 29 at 2:48
answered Jan 28 at 16:52
MefiticoMefitico
1,182218
answered Jan 28 at 16:52
MefiticoMefitico
1,182218
answered Jan 28 at 16:52
MefiticoMefitico
1,182218
1,182218
$begingroup$
The rule you mention at the beginning is actually correct but has a hypothesis that limit of $g(x) $ is non-zero.
$endgroup$
– Paramanand Singh
Jan 29 at 2:27
$begingroup$
@ParamanandSingh : Yes, but that's precisely the case in which OP tried to use the rule.
$endgroup$
– Mefitico
Jan 29 at 2:45
add a comment |
$begingroup$
The rule you mention at the beginning is actually correct but has a hypothesis that limit of $g(x) $ is non-zero.
$endgroup$
– Paramanand Singh
Jan 29 at 2:27
$begingroup$
@ParamanandSingh : Yes, but that's precisely the case in which OP tried to use the rule.
$endgroup$
– Mefitico
Jan 29 at 2:45
$begingroup$
The rule you mention at the beginning is actually correct but has a hypothesis that limit of $g(x) $ is non-zero.
$endgroup$
– Paramanand Singh
Jan 29 at 2:27
$begingroup$
The rule you mention at the beginning is actually correct but has a hypothesis that limit of $g(x) $ is non-zero.
$endgroup$
– Paramanand Singh
Jan 29 at 2:27
$begingroup$
@ParamanandSingh : Yes, but that's precisely the case in which OP tried to use the rule.
$endgroup$
– Mefitico
Jan 29 at 2:45
$begingroup$
@ParamanandSingh : Yes, but that's precisely the case in which OP tried to use the rule.
$endgroup$
– Mefitico
Jan 29 at 2:45
add a comment |
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$begingroup$
$limfrac fg=frac{lim f}{lim g}$ iff $lim f,lim g$ exist and $lim gne0$, so your second step is wrong. Say it wasn't, then your next step falters when you selectively apply the limit to $dfrac{sin x}x$ in the numerator but not to $cos x$.
$endgroup$
– Shubham Johri
Jan 28 at 16:59
$begingroup$
@ShubhamJohri: there is no faltering in second step. Convince yourself that the following is true. If $g(x) to L$ as $xto a$ then $$lim_{xto a} {f(x) pm g(x) } =lim_{xto a} f(x) pm L$$ See more details at math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Jan 29 at 2:32