Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

3 Answers 3

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

share|improve this answer
add comment

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 $0$.

share|improve this answer
add comment

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 $x \rightarrow \infty$.

share|improve this answer
1  
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
1  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.