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I would like to evaluate the following limit $$\lim_{x \rightarrow 0}\frac{\int \limits_0^{\sin x}e^{xt^{2}}dt}{\int \limits_0^{\tan x}e^{-xt^{2}}dt}$$

In order to use L'Hospital's rule I obviously need derivatives with respect to $x$. With this including x as a parameter in the upper integration limits, I can reduce the limit to $$\lim_{x \rightarrow 0}\frac{e^{x \sin^{2} x}\cos x+\int \limits_0^{\sin x}t^{2}e^{xt^{2}}dt}{e^{-x\tan^{2}x}\cos^{-2} x+\int \limits_0^{\tan x}(-t^{2})e^{-xt^{2}}dt}$$

I'm not sure if I'm approaching this in the correct way; so my questions are

  1. Am I using L'Hospital's rule correctly in the above? If so
  2. How should I proceed from now on? I feel a little stuck.

Any help is greatly appreciated.

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    $\begingroup$ Why not plug $t^2=x^2$ $\endgroup$ – Archis Welankar Mar 2 '16 at 12:03
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Yes the derivatives of the numerator and denominator are correct (e.g. from https://en.wikipedia.org/wiki/Leibniz_integral_rule). And because the integrals in the new quotient go to $0$ as $x\rightarrow 0$, you can simply plug-in $x=0\;$ and get the limit $\frac{1}{1}$

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  • $\begingroup$ Hello, many thanks for the help - the answer is indeed the one you quote; I had hoped I's used the rule correctly. Happy to accept this as an answer. Thank you. $\endgroup$ – vectorbundle Mar 2 '16 at 12:46

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