Answer
$-1$
Work Step by Step
Taylor series for $ e^x $ can be defined as: $ e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+....$
Now, $\lim\limits_{x \to \infty} x^2 (e^{1/x^2}-1)=\lim\limits_{x \to \infty} x^2 (-1+e^{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, $=\lim\limits_{x \to \infty} -1+\dfrac{1}{2x^2}-\dfrac{1}{6x^4}+....$
or, $=-1$
