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.
On
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 $$ \begin{align} e^u&=1+u+O(u^2)\\ \arctan u &=u+O(u^2). \end{align} $$
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
$$-1/2\cdot\lim_{h\to0^+}\frac{e^{-h^2}-1}{\dfrac\pi2-\arctan 1/h}$$ $$=-1/2\cdot\lim_{h\to0^+}\frac{e^{-h^2}-1}{\arctan h}$$
$$=1/2\cdot\lim_{h\to0^+}\frac{e^{-h^2}-1}{-h^2}\cdot\lim_{h\to0^+}\dfrac h{\arctan h}\cdot\lim_{h\to0^+} h=?$$