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?

This is indeterminate because $x^4\rightarrow \infty$, but $\sqrt{1+\frac{4}{x^8}}-1\rightarrow 0$.  You can multiply by the conjugate
$$
\left(\sqrt{x^8+4}-x^4\right)\left(\frac{\sqrt{x^8+4}+x^4}{\sqrt{x^8+4}+x^4}\right)=\frac{x^8+4-x^8}{\sqrt{x^8+4}+x^4}=\frac{4}{\sqrt{x^8+4}+x^4}
$$
