Evaluate $\lim_{x\to \infty} \frac{e^{\frac{1}{x^2}}-1}{2\tan^{-1} (x^2)-\pi}$ $$\lim_{x\to \infty} \frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \frac{1}{(2\tan^{-1}(x^2)-\pi)x^2}$$
$$=\lim_{x\to \infty} \frac{1}{x^2(2\tan^{-1} x^2 -\pi)}$$
How do I proceed?
 A: From here
$$\frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \frac{1}{\left(2\tan^{-1}(x^2)-\pi\right)x^2}$$
we have that by standard limit
$$\frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \to 1$$
and by $\arctan x = \frac \pi 2 -\arctan \frac1x$
$$\frac{1}{\left(2\tan^{-1}(x^2)-\pi\right)x^2}=-\frac12\frac{\frac1{x^2}}{\tan^{-1}\left(\frac1{x^2}\right)} \to -\frac12$$

To proceed by l'Hospital by $\frac1{x^2}=t \to 0$ we have
$$\lim_{x\to \infty }\frac{e^{\frac{1}{x^2}}-1}{2\tan^{-1} (x^2)-\pi}=\lim_{t\to 0 }\frac{e^{t}-1}{2\tan^{-1} \left(\frac1t\right)-\pi}=\lim_{t\to 0 }\frac{e^{t}}{-\frac2{t^2+1}}=-\frac12$$
A: Apply l'Hôpital's rule:
$$L=\lim\limits_ {x \to \infty} \dfrac {x^{-2}} {2\arctan x^2 -\pi}=\lim\limits_ {x \to \infty} \dfrac {-2x^{-3}} {2\dfrac {2x}{x^4+1}}$$
$$L==\lim\limits_ {x \to \infty} -{\dfrac {x^4}{2x^4}}=-\dfrac 12$$
A: Limit[(Exp[-(1/x^2)] - 1)/(1/x^2) 1/((ArcTan[x^2] - π) x^2), 
 x -> 0]
1/π

Limit[(Exp[-(1/x^2)] - 1)/(1/x^2) , x -> 0]
0

Limit[ (ArcTan[x^2] - \[Pi]) x^2, x -> 0]
0

So this is a quotient 0/0 and the rule of l'Hospital applies.
Series[(ArcTan[x^2] - π) x^2, {x, 0, 2}]
SeriesData[x, 0, {-Pi}, 2, 3, 1]


Series[Exp[-(1/x^2)] - 1, {x, 0, 2}]


This has to be developed as:

This give in the rule of l'Hospital:
-x^2/-πx^2 -> 1/π

So the above shown result is clear and prooved.
{Plot[{(Exp[-(1/x^2)] - 1)/(1/x^2) 1/((ArcTan[x^2] - \[Pi]) x^2), 
   N@1/\[Pi]}, {x, 0, 1}], N@1/\[Pi]}


$$\lim_{x\rightarrow\infty}\frac{(\exp\frac{1}{x^2}-1)}{2\arctan(x^2)-\pi}=\frac{1}{2}$$
The quotient is alreay "-0.5001" at "x=50".
