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: $$\dfrac{x^2+2x+3}{x^2+x+1} = 1 + \dfrac{x+2}{x^2+x+1}$$
Hence,
$$\left(\dfrac{x^2+2x+3}{x^2+x+1}\right)^x = \left(1 + \dfrac{x+2}{x^2+x+1} \right)^{\left(\dfrac{x^2+x+1}{x+2} \right) \left(\dfrac{x(x+2)}{x^2+x+1} \right)}$$
Now as $x \to \infty$, we have $\left(1 + \dfrac{x+2}{x^2+x+1} \right)^{\left(\dfrac{x^2+x+1}{x+2} \right)} \to e$ and $\left(\dfrac{x(x+2)}{x^2+x+1} \right) \to 1$.
A: $$\lim_{x \rightarrow \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$$
$$=\lim_{x \rightarrow \infty}\left(1+\frac{x+2}{x^2+x+1} \right)^x$$
$$=\lim_{x \rightarrow \infty}\left(\left(1+\frac{x+2}{x^2+x+1} \right)^\frac{x^2+x+1}{x+2}\right)^{\frac{x(x+2)}{x^2+x+1}}$$
$$=e$$  as $\lim_{x\to\infty}\frac{x(x+2)}{x^2+x+1}=\lim_{x\to\infty}\frac{(1+2/x)}{1+1/x+1/{x^2}}=1$
and $\lim_{x\to\infty}\left(1+\frac{x+2}{x^2+x+1} \right)^\frac{x^2+x+1}{x+2}=\lim_{y\to\infty}\left(1+\frac1y\right)^y=e$
A: 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 $L=e$.
A: $$
\begin{aligned}
\lim_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x
& = \lim _{x\:\to \infty }\left(e^{x\ln\left(\frac{x^2+2x+3}{x^2+x+1}\right)}\right)
\\& \approx \lim _{x\:\to \infty }\left(e^{x\left(\frac{x^2+2x+3}{x^2+x+1}-1\right)}\right)
\\& = \lim _{x\:\to \infty }\left(e^{\frac{x^2+2x}{x^2+x+1}}\right)
\\& = \color{red}{e}
\end{aligned}
$$
