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?

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
$$\lim_{x\to \infty }\frac{e^{\frac{1}{x^2}}-1}{2\tan^{-1} (x^2)-\pi}=\lim_{t\to 0 }\frac{e^{t}-1}{2\tan^{-1} \left(\frac1t\right)-\pi}=\lim_{t\to 0 }\frac{e^{t}}{-\frac2{t^2+1}}=-\frac12$$
