If $f(x)=\left(\frac{x^2+5 x+3}{x^2+x+2}\right)^x$, then $\lim\limits_{x \rightarrow \infty} f(x)$ is

$e^4$

$f(x)=\lim\limits_{x \rightarrow \infty}\left(\frac{x^2+5 x+3}{x^2+x+2}\right)^x$

$\log f(x)=\lim\limits_{x \rightarrow \infty} x . \log \left(\frac{x^2+5 x+3}{x^2+x+2}\right)$

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

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

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

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

$=\lim\limits_{x \rightarrow \infty}\left(\frac{x^2\left(4+\frac{1}{x}\right)}{x^2\left(1+\frac{1}{x}+\frac{2}{x}\right)}\right)=4 \Rightarrow f(x)=e^4$

Hence (1) is the correct answer.