Value of $\lim_{n\to\infty}{(1+\frac{2n^2+\cos{n}}{n^3+n})^n}$
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, $$ \mathop {\lim }\limits_{n \to \infty } \bigg(1 + \frac{{2n^2 + \cos n}}{{n^3 + n}}\bigg)^n = e^2 . $$
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,
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.