Answer
$-1$
Work Step by Step
$\lim\limits_{x \to \infty} x^2 (e^{1/x^2}-1)=\lim\limits_{x \to \infty} x^2 (-1+e^{(\frac{1}{x^2})}) $
or, $=\lim\limits_{x \to \infty} x^2 [-1+1-\dfrac{1}{x^2}+\dfrac{1}{2x^4}-\dfrac{1}{6x^6}+.....]$
or, $=\lim\limits_{x \to \infty} (-1) + \lim\limits_{x \to \infty} \dfrac{1}{2x^2}-\lim\limits_{x \to \infty} \dfrac{1}{6x^4}+....$
So, $\lim\limits_{x \to \infty} x^2 (e^{1/x^2}-1)=-1$
