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$$\displaystyle\lim_{x\to0}\left(\frac{1}{x^5}\int_0^xe^{-t^2}\,dt-\frac{1}{x^4}+\frac{1}{3x^2}\right)$$

I have this limit to be calculated. Since the first term takes the form $\frac 00$, I apply the L'Hospital rule. But after that all the terms are taking the form $\frac 10$. So, according to me the limit is $ ∞$. But in my book it is given 1/10. How should I solve it?

Olivier Oloa
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Debarun Mukherjee
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  • The limit of just the first term is infinity. Are you getting something different when you use l'Hôpital on the first term? If the answer is truly $1/10$ then these three terms need to work together to make it happen since each one by itself has an infinite limit. –  Aug 26 '16 at 17:30
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    @Debarun Mukherjee Just use L'hospital's rule. For the numerator use Leibnitz rule. After applying L'hospital 3times you will get something like e^x/10 now put x=0 and here's the answer – Sujan Dutta Aug 26 '16 at 17:46

3 Answers3

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One may recall that, as $t \to 0$, by the Taylor series expansion $$ e^{-t^2}=1-t^2+\frac{t^4}2+O(t^6) $$ giving, as $x \to 0$, $$ \int_0^xe^{-t^2}dt=x-\frac{x^3}3+\frac{x^5}{10}+O(x^7) $$ and, as $x \to 0$,

$$ \frac1{x^5}\int_0^xe^{-t^2}dt-\frac1{x^4}+\frac1{3x^2}=\frac1{10}+O(x^2) $$

from which one deduces the desired limit.

Olivier Oloa
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By bringing the fractions to the same denominator, start by writing the limit as $$\displaystyle\lim_{x\to0}\frac{3\int_0^xe^{-t^2}\,dt -3x+x^3 }{3x^5}$$

Now, since this is of the form $0/0$ by L'H and FTC you get $$\displaystyle\lim_{x\to0}\frac{3e^{-x^2}-3+x^2 }{15x^4}$$

From here it is easy.

N. S.
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  • Doesn't this answer imply the limit is $\infty$? I don't see where the $1/10$th comes from here. – Alex Ortiz Aug 26 '16 at 17:42
  • @AOrtiz Why infinity? This limit is $\frac{3-3}{0}$ which is still undetermined ;) – N. S. Aug 26 '16 at 17:43
  • So what is actually happening? Is the limit $1/10$, or is it undetermined? Your answer claims the rest is easy, but the OP suggests the answer is something yours is not. – Alex Ortiz Aug 26 '16 at 17:45
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    @AOrtiz Undetermined/Indefinite means you have to do more computations. Since it is of the form $0/0$ you can use L'H again. And since the denominator is $x^4$ you need to use L'H most likely 4 more times. – N. S. Aug 26 '16 at 17:47
  • @N. S. yes I get it. My mistake was that I applied the L'H rule on the left most term without first simplifying the expression. – Debarun Mukherjee Aug 26 '16 at 17:53
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Hint: $e^{-u} = 1-u+u^2/2 + O(u^3).$

zhw.
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    @SujanDutta You should point that out to the OP. – StubbornAtom Aug 26 '16 at 17:41
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    I do not like L'Hopital in general, because it tends to be a mechanincal computation that doesn't yield much insight. (Although LHR does have its uses.) Using the hint gives a quicker proof and it also shows where the problem came from. – zhw. Aug 26 '16 at 17:49