How to prove the limit of a sequence by definition: $\lim((n^2 + 1)^{1/8} - n^{1/4}) = 0$? I have to prove the limit of a sequence using $\epsilon - N$ definition: $$\lim_{n\to\infty}((n^2 + 1)^{1/8} - n^{1/4}) = 0$$. 
Attempt:
We want to show: $\forall \epsilon>0$ $\exists N \in \mathbb{N}$, s.t. if $n \ge N$, then $|(n^2 + 1)^{1/8} - n^{1/4} - 0|<\epsilon$. So we need to find N as a function of $\epsilon$, s.t. N > $f(\epsilon)$.
$(n^2 + 1)^{1/8} - n^{1/4}$ = $((n^2 + 1)^{1/8} - n^{1/4}) * \frac{(n^2 + 1)^{1/8} + n^{1/4}}{(n^2 + 1)^{1/8} + n^{1/4}}$ = $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}}$. 
$\frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}} < \frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}}$, so any n, solving $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}} < \epsilon$, will suffice. 
Then $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}}$ = $(n + \frac{1}{n})^{1/4} - n^{1/4}$.
I do not know how to proceed further. I tried the same trick with multiplying the last equation by $\frac{(n + \frac{1}{n})^{1/4} + n^{1/4}}{(n + \frac{1}{n})^{1/4} + n^{1/4}}$, but I did not get anything meaningful. 

Our expression equals $(n^2+1)^{1/8} - (n^2)^{1/8}.$ By the mean value theorem,
$$\tag 1 (n^2+1)^{1/8} - (n^2)^{1/8} =(1/8)c_n^{-7/8}\cdot 1,$$
where $c_n\in (n^2, n^2+1).$ It follows that $(1) \le (1/8)(n^2)^{-7/8} = (1/8)n^{-7/4}.$ You're now set up for an $\epsilon$-$N$ proof.
