Converting limit to e exponent form I don't understand why this is true. I would like to see the process of converting this limit to e,  to better understand this topic.
$$\lim_{n \to \infty }\left({n^{2}+n \over n^{2}+1}\right)^{n} = e^1 = e$$ 

Let $y=\lim_{x\to\infty}(1+\frac{1}{x})^x$ then:
$$\begin{align*}\ln y &= \ln \space[\lim_{x \to \infty} (1+\frac{1}{x})^x] \\ &= \lim_{x \to \infty}[\ln (1+\frac{1}{x})^x] \\ &= \lim_{x \to \infty}[x\ln (1+\frac{1}{x})] \\ &= \lim_{x \to \infty}[ \frac{\ln (1+\frac{1}{x})}{\frac{1}{x}}]\end{align*}$$
L'Hopital's rule is now used because other wise the limit would result in $\frac{0}{0}$:
$$ \begin{align*} \ln y &= \lim_{x \to \infty}[\frac{(\frac{\frac{-1}{x^2}}{1+\frac{1}{x}})}{\frac{-1}{x^2}}] \\ &= \lim_{x \to \infty} 1+\frac{1}{x} \\ &= 1\end{align*}$$
We now have $\ln y = 1$, this implies $y=e$. This concludes the proof. 
