How do I find $\lim_{n\to \infty}\left [\left (1+\frac{2}{n^a}\right )^{-n^b}n^c\right ]$ for real $a,b,c$ and $n\geq 1$? 
Evaluate $\displaystyle\lim_{n\to \infty}\left [\left (1+\frac{2}{n^a}\right )^{-n^b}n^c\right ]$ for real $a,b,c$ and $n\geq 1$.

I am not sure how to do this but the hint given is to consider various cases of $a>b$, $a<b$, $a=b$ and $c>0$, $c<0$.
 A: $$t=\displaystyle\lim_{n\to \infty}\left (1+\frac{2}{n^a}\right )^{-n^b}n^c=\large \displaystyle\lim_{n\to \infty} \frac{n^c}{e^\frac{2n^b}{n^a}}$$. 


*

*If $b>a$ then $t=0$

*If $b< a$ then $t\to\infty$

*If $a=b$ and $c<0$ then $t=0$ but If $a=b$ and $c>0$ then $t\to\infty$

A: Standard limit rule says$$\lim_{n\to \infty}\left [\left (1+\frac{2}{n^a}\right )^{-n^b}\right]=e^{-2n^{b-a}}$$ 
So result is $$\lim\limits_{n\to \infty}{n^c\over e^{2n^{b-a}}}$$
$b>a\implies\lim\limits_{n\to \infty}{n^c\over e^{2n^{b-a}}}\to 0 \ \forall $ finite $c$  
$b=a\implies \lim\limits_{n\to \infty}{n^c\over e^{2n^{b-a}}} \to \infty\ \ $ if $c\gt 0$, $1$ if $c=0$ and $0$ if $c\lt 0$ 
$b<a \implies \lim\limits_{n\to \infty}{n^c\over e^{2n^{b-a}}} \to \infty$
A: $$=\frac{n^c}{(1+\frac{2}{n^a})^{n^b}}$$
now let $$l=(1+\frac{2}{n^a})^{n^b}$$
$$\ln l={n^b}\ln({1+\frac{2}{n^a}})=\frac{2n^b}{n^a}\frac{\ln({1+\frac{2}{n^a}})}{2/n^{a}}=\frac{2n^b}{n^a}.1$$
$$\therefore l=e^ {\frac{2n^b}{n^a}}$$
therefore original limit is $$=\frac{n^c}{e^{2n^{b-a}}}$$ 
now exponential function raises a lot faster than a finite polynomial thus limit would $\infty$ if $b<=a$ and $0$ if $b>a$ assuming $c$ is positive 
