1
$\begingroup$

I wish to compute the following limit

$$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})$$

Wolfram Alpha gives the answer as $1/2$. I've tried L'Hopital's rule, but I can't seem to get it to work.

$\endgroup$
2
$\begingroup$

Try

$$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})= \lim_{n \rightarrow \infty} \sqrt{n+1}\frac{n-1}{\sqrt{n^3 + n} + \sqrt{n^3 + 1}}=\lim_{n \rightarrow \infty} \sqrt{1+1/n}\frac{1-1/n}{\sqrt{1 + 1/n^2} + \sqrt{1 + 1/n^3}}=1/2$$

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

$$ \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1}) \cdot \frac{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}}{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}} = \frac{\sqrt{n+1}(n-1)}{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}}$$ Now divide numerator and denominator by $\sqrt{n^3}.$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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