Calculate the limit: $\lim_{n\to\infty}n^2\left(\left(1+\frac{1}{n}\right)^8-\left(1+\frac{2}{n}\right)^4\right)$ Calculate the limit:
$\lim\limits_{n\to\infty}n^2\left(\left(1+\dfrac{1}{n}\right)^8-\left(1+\dfrac{2}{n}\right)^4\right)$
My first suggestion was that $\lim\limits_{n\to\infty} = 0$. As in both brackets as ${n\to\infty}$: $\dfrac{1}{n}$ and $\dfrac{2}{n}$ will ${\to0}$, so it was going to be $\lim\limits_{n\to\infty}n^2\left(\left(1\right)^8-\left(1\right)^4\right) => \lim\limits_{n\to\infty} = 0$.
My second suggestion was using the properties of $\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n = e$ and $\lim\limits_{n\to\infty}\left(1+\dfrac{k}{n}\right)^n = e^k$ to find limits of enclosing brackets expressions:
1. $\lim\limits_{n\to\infty}(1+\dfrac{1}{n})^8 => \lim\limits_{n\to\infty}((1+\dfrac{1}{n})^n)^{\frac{8}{n}} => e^{\frac{8}{n}}$
2. $\lim\limits_{n\to\infty}(1+\dfrac{2}{n})^4 => \lim\limits_{n\to\infty}((1+\dfrac{2}{n})^n)^{\frac{4}{n}} => (e^2)^{\frac{4}{n}} => e^{\frac{8}{n}}$
It brought me again to $\lim\limits_{n\to\infty}n^2 *(e^{\frac{8}{n}} - e^{\frac{8}{n}}) => \lim\limits_{n\to\infty} = 0$
However, $0$ is a wrong answer. How to find the limit?  
P.S. I am self-study calculus newbie, so please answer as easy as possible (don't know L'Hôpital's rule yet).
 A: Let $x=1+\frac{1}{n}$ and  $y=1+\frac{2}{n}$ then 
$$x^8-y^4=(x^2-y)(x^2+y)(x^4+y^2)$$ and 
$$x^2-y=\frac{1}{n^2}$$
Thus the limit is 
$$(x^2+y)(x^4+y^2) \to 4$$
A: Set $1/n=h$ to get $$\lim_{h\to0}\dfrac{(1+h)^8-(1+2h)^4}{h^2}$$
Now $\displaystyle(1+2h)^4=1+\binom41(2h)+\binom42(2h)^2+O(h^3)$
and $\displaystyle(1+h)^8=1+8h+\binom82h^2+O(h^3)$
A: You have the product of two terms, one of which goes to $\infty$ and one of which goes to $0$.  You cannot conclude that the product goes to $0.$
As to how to do the problem, you can expand out $\left(1+\frac{1}{n}\right)^8$ and $\left(1+\frac{1}{2n}\right)^4$ by the binomial theorem, multiply by $n^2$ and compute the limit.
The trick is that you don't have to go past the second degree terms, because once you have a power of $3$ or greater in the denominator, the product will go to $0$.  For example, $$
\left(1+\frac{1}{n}\right)^8=1+\frac{8}{n}+\frac{28}{n^2}+o\left(\frac{1}{n^2}\right)$$ where by $o\left(\frac{1}{n^2}\right)$ I just mean terms that are very small compared to $1/n^2.$  Do the same thing with $\left(1+\frac{1}{2n}\right)^4$ 
Can you take it from here? 
