If ${\lim }_{x\to \infty }(1+\frac{a}{x}+\frac{b}{x^2}{)}^{2x}=e^2,$ then the values of $a$ and $b,$ are

The limit is of the form $1^{\infty }.$ We use the formula ${\lim }_{x\to c}f(x{)}^{g(x)}=e^{{\lim }_{x\to c}g(x)(f(x)-1)}.$${\lim }_{x\to \infty }(1+\frac{a}{x}+\frac{b}{x^2}{)}^{2x}=e^{{\lim }_{x\to \infty }2x(1+\frac{a}{x}+\frac{b}{x^2}-1)}$.
$=e^{{\lim }_{x\to \infty }2x(\frac{a}{x}+\frac{b}{x^2})}$
$=e^{{\lim }_{x\to \infty }(2a+\frac{2b}{x})}$
$=e^{2a}$.
Given the limit equals $e^2$:
$e^{2a}=e^2\Rightarrow 2a=2\Rightarrow a=1$.
The value of $b$ does not affect the limit as its term $\frac{2b}{x}$ tends to $0$ regardless of $b$ 's value.
Thus $a=1$ and $b\in \mathbb{R}$.