Value of $\lim_{n\to\infty}{(1+\frac{2n^2+\cos{n}}{n^3+n})^n}$ How should one go about computing $$\lim_{n\to\infty}{\left(1+\frac{2n^2+\cos{n}}{n^3+n}\right)^n}\quad?$$ What surprised me about this is that 
$$\lim_{n\to\infty}{\left(1+\frac{2n^2+\cos{n}}{n^3+n}\right)^\frac{n^3+n}{2n^2+\cos{n}}}=1$$(according to wolfram), instead of $e$, which is what I expected. Could someone comment on that also?
 A: Hint: Show that
$$
\lim _{n \to \infty } n\ln \bigg(1 + \frac{{2n^2  + \cos n}}{{n^3  + n}}\bigg) = 2.
$$
Elaborating: 
$$
 n\ln \bigg(1 + \frac{{2n^2  + \cos n}}{{n^3  + n}}\bigg) = n\ln \bigg(1 + \frac{{2 + \cos (n)/n^2 }}{{n + 1/n}}\bigg) = na_n \frac{{\ln (1 + a_n )}}{{a_n }},
$$
where
$$
a_n = \frac{{2 + \cos (n)/n{}^2}}{{n + 1/n}}.
$$
Noting that $a_n \to 0$ as $n \to \infty$ and 
$$
\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{1/(1 + x)}}{1} = 1,
$$
we conclude that
$$
\lim _{n \to \infty } \frac{{\ln (1 + a_n )}}{{a_n }} = 1
$$
and, in turn,
$$
\lim _{n \to \infty } n\ln \bigg(1 + \frac{{2n^2  + \cos n}}{{n^3  + n}}\bigg) = \lim _{n \to \infty } na_n  = \lim _{n \to \infty } n\frac{{2 + \cos (n)/n^2 }}{{n + 1/n}} = 2.
$$
Thus,
$$
\mathop {\lim }\limits_{n \to \infty } \bigg(1 + \frac{{2n^2  + \cos n}}{{n^3  + n}}\bigg)^n  = e^2 .
$$
A: Lets consider the more general case of an arbitrary function.  Then we have the following theorem:
Theorem: Given a function $g(n)$ such that $g_0=\lim_{n\rightarrow \infty} g(n)$ exists, we have $$\lim_{n\rightarrow \infty }\left(1+\frac{g(n)}{n}\right)^n=e^{g_0}.$$ 
Example: In particular, for your above question, $g(n)=\frac{2n^2+\cos(n)}{n^2+1}$, so that $g_0 =2$, and hence the value of the original limit is $e^2$.
Proof of theorem:  Let $f(n)=\frac{g(n)}{g_0}$ so that  $f(n)=1+o(1)$.  Then since $\lim_{n\rightarrow \infty} \left(1+\frac{a}{n}\right)^n=e^a$ we see that $$\lim_{n\rightarrow \infty} \left(1+\frac{g(n)}{n}\right)^\frac{n}{f(n)}=e^{g_0}.$$  Then, because  $f(n)=1+o(1)$, it follows that $\frac{n}{f(n)}=n+o(n)$.  But, $$\lim_{n\rightarrow \infty} \left(1+\frac{O(1)}{n}\right)^{o(n)}=1,$$ so we conclude that $$\lim_{n\rightarrow \infty }\left(1+\frac{g(n)}{n}\right)^n=e^{g_0}.$$
I hope that helps,
A: What you should use is that 
$$ \lim_{x \to 0} (1+x)^{\frac{1}{x}}=e$$
Then, if you have to calculate $\lim_{n \to \infty} x_n^{y_n}$ where $x_n \to 1$ and $y_n \to \infty$ you proceed as follows:


*

*Denote $a_n=x_n-1$ so $a_n \to 0$. 

*Now you have to calculate $\lim_{n \to \infty} (1+a_n)^{y_n}$.

*Transform the exponent so that you get the $e$-limit presented above:
$$ \lim_{n \to \infty} ((1+a_n)^{\frac{1}{a_n}})^{a_n y_n} =e^L$$
where $\displaystyle L=\lim_{n \to \infty}a_n y_n$, which usually is pretty simple to calculate.
