If $f(x)=\left(\frac{x}{2+x}\right)^{2 x}$, then $\lim\limits_{x \rightarrow \infty} f(x)$ is

$e^{-4}$

$f(x)=\left(\frac{x}{x+2}\right)^{2 x}$

$\log f(x)=\lim\limits_{x \rightarrow \infty} 2 x \log \left(\frac{x+2-2}{x+2}\right)=2 x\left(\log \left(1-\frac{2}{x+2}\right)\right)$

$=\lim\limits_{x \rightarrow \infty}-1 \times 2 x\left(\frac{2}{x+2}+\frac{1}{2}\left(\frac{2}{x+2}\right)^2+\frac{1}{3}\left(\frac{x}{x+2}\right)^3 .....\right)$

$=\lim\limits_{x \rightarrow \infty}-2 x \frac{2}{x+2}\left(1+\frac{1}{x+2}+.....\right)=\lim\limits_{x \rightarrow \infty} \frac{-4}{1+\frac{2}{x}}=-4$

$f(x)=e^{-4}$

Hence (1) is the correct answer.