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.
$$\lim_{n \to \infty }\left({n^{2}+n \over n^{2}+1}\right)^{n} = e^1 = e$$ 
 A: 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. 
A: Are you familiar with $$\lim_{n\to\infty}\left(1+{1\over n}\right)^n=e$$ If so, do you see how to compare the limit you want with the one you know?
A: Note that
$$({n^{2}+n \over n^{2}+1})^{n}=({1+\frac{1}{n} \over 1+\frac{1}{n^{2}}})^n=\frac{(1+\frac{1}{n})^n}{(1+\frac{1}{n^{2}})^n}.$$
Note also that
$$\lim_{n \to \infty }(1+\frac{1}{n})^n=e.$$
On the other hand, 
$$\lim_{n \to \infty }(1+\frac{1}{n^{2}})^n=\lim_{n \to \infty }e^{n\ln(1+\frac{1}{n^{2}})}
=e^{\lim_{n \to \infty }\frac{\ln(1+\frac{1}{n^{2}})}{\frac{1}{n}}}.$$
By L'Hosiptal Rule, 
$$\lim_{n \to \infty }\frac{\ln(1+\frac{1}{n^{2}})}{\frac{1}{n}}=\lim_{n \to \infty }\frac{(\frac{1}{1+\frac{1}{n^{2}}})(-\frac{2}{n^3})}{(-\frac{1}{n^2})}=0.$$
Combining the above the equalities, we have
$$\lim_{n \to \infty }(1+\frac{1}{n^{2}})^n=e^0=1.$$
Therefore, 
$$\lim_{n \to \infty }({n^{2}+n \over n^{2}+1})^{n} = \frac{\lim_{n \to \infty }(1+\frac{1}{n})^n}{\lim_{n \to \infty }(1+\frac{1}{n^{2}})^n}=\frac{e}{1}=e.$$
A: Another way to show that
$\lim_{n \to \infty }(1+\frac{1}{n^{2}})^n=1
$
is to use this "contra-Bernoulli-inequality"
(as I call it):
If 
$n$ is a positive integer and
$ 0 \le  x < 1/n$
then
$(1+x)^n \le 1/(1-x n)$.
This is readily proved by induction, writing it in the form
$(1-x n)(1+x)^n \le 1$.
Setting $x = 1/n^2$, we get
$$(1+1/n^2)^n \le 1/(1-1/n)
= 1+\frac{1}{n-1}.
$$
