Help in evaluating limit of the given function. 
Evaluate the limit $$\lim_{x\to \infty} \left(\cfrac{x^2+5x+3}{x^2+x+2}\right)^x $$

I'm not sure how to evaluate this limit. This is what I've done yet: 
$$\begin{align} 
\lim_{x\to \infty} \left\{1 + \left(\cfrac{x^2+5x+3}{x^2+x+2} - 1\right) \right\} ^x\\ 
=\ \lim_{x\to \infty} \left\{1 + \left(\cfrac{4x+1}{x^2+x+2}\right) \right\}^x\end{align}$$
Not sure where to proceed from here. Out of ideas! 
Any help will be greatly appreciated.

$$\begin{align}\lim_{x\to\infty}\left(\frac{x^2+5x+3}{x^2+x+2}\right)^x&=\lim_{x\to\infty}\left(1+\frac{1}{\frac{x^2+x+2}{4x+1}}\right)^{\frac{x^2+x+2}{4x+1}\cdot\frac{x(4x+1)}{x^2+x+2}}\\&=\lim_{x\to\infty}\left(\left(1+\frac{1}{\frac{x^2+x+2}{4x+1}}\right)^{\frac{x^2+x+2}{4x+1}}\right)^{\frac{4+\frac 1x}{1+\frac 1x+\frac{2}{x^2}}}\\&=e^4\end{align}$$
Here, note that $\lim_{x\to\infty}\frac{x^2+x+2}{4x+1}=+\infty$ and that $\lim_{y\to\infty}\left(1+\frac 1y\right)^y=e$.
