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I solved the following limit using L'Hospital's rule, but can't seem to solve it without using L'Hospital's. $$\lim_{x\to\infty} \frac{e^{-1/x^2}-1}{2\arctan x-\pi}$$

I would like a hint as to how to get started.

I was also wondering how to approach inverse trigonometric functions in general when they appear in limits, since I didn't understand any solutions to this type of problem that I looked up.

Olivier Oloa
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Ankita
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2 Answers2

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As $\arctan y+\operatorname{arccot} y=\dfrac\pi2$

Set $1/x=h$ and use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function? to get

$$-1/2\cdot\lim_{h\to0^+}\frac{e^{-h^2}-1}{\dfrac\pi2-\arctan 1/h}$$ $$=-1/2\cdot\lim_{h\to0^+}\frac{e^{-h^2}-1}{\arctan h}$$

$$=1/2\cdot\lim_{h\to0^+}\frac{e^{-h^2}-1}{-h^2}\cdot\lim_{h\to0^+}\dfrac h{\arctan h}\cdot\lim_{h\to0^+} h=?$$

Community
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lab bhattacharjee
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You may observe that, as $x \to +\infty$, we have $$ \frac{e^{-1/x^2}-1}{2\arctan x-\pi}=\frac{(1-1/x^2)-1+O(1/x^4)}{2(\pi/2-\arctan (1/x))-\pi}=\frac{-1/x^2+O(1/x^4)}{-1/x+O(1/x^2)}=1/x+O(1/x)\to 0 $$ where we have used, as $u \to 0$, the standard Taylor series expansions $$ \begin{align} e^u&=1+u+O(u^2)\\ \arctan u &=u+O(u^2). \end{align} $$

Olivier Oloa
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  • I am a high school student and have not been taught (at least in my country) about this 'O' thing you're using, so please explain otherwise. – Ankita Apr 27 '15 at 13:08
  • It is a powerful tool... Another way to see this is, as $u \to 0$: $$ \frac{e^u-1}u = 1+\epsilon(u)$$ $$ \frac{\arctan u-u}u = \epsilon(u)$$ with $\epsilon(u) \to 0.$Thanks! – Olivier Oloa Apr 27 '15 at 13:12
  • Thank you, but i don't understand this epsilon thing either. I probably should be visiting a dumber forum. – Ankita Apr 28 '15 at 17:05