Limit of $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$ I have to determine the following:
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)=\lim\limits_{x \rightarrow \infty}(\sqrt{x^8(1+\frac{4}{x^8})}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4\sqrt{1+\frac{4}{x^8}}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4(\sqrt{1+\frac{4}{x^8}}-1)= \infty$
Could somebody please check, if my solution is correct?

A short way to (non-rigorously) find the limit is to observe that for large $x$, 
$$
\sqrt{x^8+4} \approx \sqrt{x^8}=x^4
$$
so that for large $x$ (especially in $\lim_{x \to \infty}$)
$$
\sqrt{x^8+4}-x^4 \approx x^4-x^4=0
$$
So the limit must be $0$. 
