calculus - Calculate: $limlimits_{x to infty}left(frac{x^2+2x+3}{x^2+x+1} right)^x$

How do I calculate the following limit without using l'Hôpital's rule?

$$\lim_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$$

Answer

$$\lim_{x \rightarrow \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$$

$$=\lim_{x \rightarrow \infty}\left(1+\frac{x+2}{x^2+x+1} \right)^x$$

$$=\lim_{x \rightarrow \infty}\left(\left(1+\frac{x+2}{x^2+x+1} \right)^\frac{x^2+x+1}{x+2}\right)^{\frac{x(x+2)}{x^2+x+1}}$$

$$=e$$ 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 $\lim_{x\to\infty}\left(1+\frac{x+2}{x^2+x+1} \right)^\frac{x^2+x+1}{x+2}=\lim_{y\to\infty}\left(1+\frac1y\right)^y=e$