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How do I find this limit: $$I=\lim_{x\to 0}\dfrac{\displaystyle\int_{x}^{x^2}e^{x\sqrt{1-y^2}}dy}{\arctan{x}}$$

My try: $$I=\lim_{x\to 0}\dfrac{\displaystyle\int_{x}^{x^2}e^{x\sqrt{1-y^2}}dy}{x}$$ so $$I=\lim_{x\to 0}\left(2xe^{x\sqrt{1-x^4}}-e^{x\sqrt{1-x^2}}+\int_{x}^{x^2}e^{x\sqrt{1-y^2}}\cdot\sqrt{1-y^2}dy\right)=-1$$

I think this problem have other nice methods,Thank you

I guess this problem can use Squeeze theorem:http://en.wikipedia.org/wiki/Squeeze_theorem

I konw this $$e^x>1+x$$

math110
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  • Offcourse you can use L'hopital's rule. Use Leibniz rule for the numerator. – Mhenni Benghorbal Dec 09 '13 at 02:54
  • Try to avoid using \displaystyle in the title of the question, since it appears too big on the main page. –  Dec 09 '13 at 02:54
  • If you edit your post, do not remove what you wrote initially. Initially, you had that you wanted to prove it using L'Hospital rule. –  Dec 09 '13 at 03:58

2 Answers2

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You can use L'Hospital's rule. All you need is a disguised version of the fundamental theorem of calculus, which is called Leibniz integral rule and reads as shown below: $$\dfrac{d}{dx}\left(\int_{f(x)}^{g(x)} h(x,y)dy\right) = \int_{f(x)}^{g(x)} \dfrac{\partial h(x,y)}{\partial x}dy + h(x,g(x)) g'(x) - h(x,f(x))f'(x)$$ Use this to differentiate the numerator and see what happens.

  • Hello,I know this ,But this problem can without this methods? Thank you – math110 Dec 09 '13 at 02:53
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    @chinamath You said you wanted to use L'Hospital rule. If you want to use it, then you need to differentiate the numerator, in which case, you need to use Leibniz rule. –  Dec 09 '13 at 02:54
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I give you here a very lazy way for solving the problem : expand the exponential as a Taylor series (two terms will be sufficient), compute the antiderivative and the integral. The result of this last step is
-(1/6) Exp[x] x (6 - 6 x - x^3 + x^6)
Develop this result again as a Taylor series around x= 0; you obtain
-x + x^3/2 + x^4/2 + (7 x^5)/24
Since the Taylor series of ArcTan[x] is x - x^3/3, then your limit is -1

Claude Leibovici
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