limit to Singularity

It is of the form $1^{\infty}$

so

$e^{\large {\displaystyle \lim_{x\to 0} (1+e^{\frac{-1}{x^{2}}}tan^{-1}\frac{1}{x^{2}}+xe^{\frac{-1}{x^{2}}}sin\frac{1}{x^{4}}-1)e^{\frac{1}{x^{2}}}}}$

$e^{\large {\displaystyle \lim_{x\to 0} (e^{\frac{-1}{x^{2}}}tan^{-1}\frac{1}{x^{2}}+xe^{\frac{-1}{x^{2}}}sin\frac{1}{x^{4}})e^{\frac{1}{x^{2}}}}}$

cancel out term $\large{e^{\frac{1}{x^{2}}}}$

we are left with

$e^{\large {\displaystyle \lim_{x\to 0} (tan^{-1}\frac{1}{x^{2}}+xsin\frac{1}{x^{4}})}}$

now $\large {\displaystyle \lim_{x\to 0} tan^{-1}\frac{1}{x^{2}}\to \frac{\pi}{2}}$

and $\large {\displaystyle \lim_{x\to 0} xsin\frac{1}{x^{4}} \to 0}$

becoz it can be interpreted as $0\times value~between ~-1~and~1$ which is zero.

hence the answer is

$\huge{\boxed{e^{\frac{\pi}{2}}=4.81}}$