Limit of $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$

Question

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?

Answer 1

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

Answer 2

We have $$ x^4 (\sqrt{1 + \frac{4}{x^8}}-1)= x^4 (\frac{2}{x^8} + O(x^{-8})) =\frac{2}{x^4} + O (x^{-4}) \rightarrow 0 $$ when $x \rightarrow \infty$.

Answer 3

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 $0$.