# 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?

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This is a good start except for the final conclusion. When you take the limit of a product and one term goes to zero and the other to infinity, you usually need to convert it to a quotient so you can use L'Hopital's rule. – aschepler Dec 8 '13 at 23:53
$$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)=(\sqrt{x^8+4}-x^4)\lim\limits_{n \rightarrow \infty}(1)=\sqrt{x^8+4}-x^4$$Now seriously, the last equality is wrong, you have $+\infty\times 0$. – Git Gud Dec 8 '13 at 23:53
L'Hopital's rule is unknown for us so far – fear.xD Dec 8 '13 at 23:56

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

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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 $0$.

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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 $x \rightarrow \infty$.

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Are you familiar with Landau "big-O" notation? It might be useful in making this answer a bit more rigorous. – robjohn Dec 9 '13 at 0:03
@Ti-Kong ng : yes, your answer doesn't look like it makes sense the way you wrote it (I did not downvote). – Stefan Smith Dec 9 '13 at 0:05
Sorry, i rewrite it in a way more rigorous. – Tī-Kong n̂g Dec 9 '13 at 0:09