Calculate: $\lim\limits_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$ How do I calculate the following limit without using l'Hôpital's rule?
$$\lim_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$$

A slightly different take.  Let $L$ be the limit in question.  Then we have
$$\begin{align}\log{L} &= \lim_{x \rightarrow \infty}x \log{\left(\frac{x^2+2 x+3}{x^2+x+1}\right)}\\ &= \lim_{x \rightarrow \infty}x \log{\left( 1+\frac{x+2}{x^2+x+1}\right)}\\ &= \lim_{x \rightarrow \infty} \frac{x(x+2)}{x^2+x+1}\end{align}$$
Therefore $\log{L}=1$ and $L=e$.
