<s>[INST] 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?
 [/INST] Apply l'Hôpital's rule:
$$L=\lim\limits_ {x \to \infty} \dfrac {x^{-2}} {2\arctan x^2 -\pi}=\lim\limits_ {x \to \infty} \dfrac {-2x^{-3}} {2\dfrac {2x}{x^4+1}}$$
$$L==\lim\limits_ {x \to \infty} -{\dfrac {x^4}{2x^4}}=-\dfrac 12$$
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