<s>[INST] 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. 
 [/INST] Just keep (fearlessly) your argument going:
$$(n^2 + 1)^{1/8} - n^{1/4} = \frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}}$$
consequently
\begin{align*}(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}}\cdot \frac{(n^2 + 1)^{1/4} + n^{1/2}}{(n^2 + 1)^{1/4} + n^{1/2}} \\ & =\frac{(n^2 + 1)^{1/2} - n}{((n^2 + 1)^{1/8} + n^{1/4})\cdot ((n^2 + 1)^{1/4} + n^{1/2})}\end{align*}
and again, doing the same trick
$$(n^2 + 1)^{1/8} - n^{1/4} = \frac{(n^2 + 1) - n^2}{((n^2 + 1)^{1/8} + n^{1/4})\cdot ((n^2 + 1)^{1/4} + n^{1/2})((n^2 + 1)^{1/2} + n)}.$$
The $n^2$ now cancel and you get
$$(n^2 + 1)^{1/8} - n^{1/4} = \frac{1}{((n^2 + 1)^{1/8} + n^{1/4})\cdot ((n^2 + 1)^{1/4} + n^{1/2})((n^2 + 1)^{1/2} + n)}.$$
Now finding your $N$ is easy.
Remark: this is obviously a half joke/half overkill, but works nonetheless.
</s>