# Limits with squeeze theorem: $\lim_{n\to\infty}\; \sqrt{n}\, \left(\sqrt{n^3+n} - \sqrt{n^3+1}\right)$

I can't find a way to solving this limit I tried using the squeeze theorem but i don't get to anywhere :

The question is to find the limit:

$$\lim_{n\to\infty}\; \sqrt{n}\, \left(\sqrt{n^3+n} - \sqrt{n^3+1}\right)$$

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Welcome to M.SE, please use MathJax for equations (we edited it for you this time). You can insert them with the dollar-sign (), or the double dollar-sign – AlexR Aug 23 '13 at 11:30 ## 1 Answer Hints: We can multiply numerator and denominator by the conjugate of the given expression: $$\sqrt a - \sqrt b = (\sqrt a - \sqrt b)\cdot\frac{\sqrt a + \sqrt b}{\sqrt a + \sqrt b} = \frac {a - b}{\sqrt a + \sqrt b}$$ That gives us: \begin{align} \lim_{n\to \infty}\sqrt n(\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}} & = \lim_{n\to\infty} \frac{\sqrt n(n - 1)}{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}}\\ \\ \end{align} Divide through byn^{3/2}\$

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Very clear explanation +1 –  Amzoti Aug 24 '13 at 0:41
@amWhy: +1 for your hint (where are you? are you OK?) –  Software Aug 24 '13 at 8:49
@Software Here I am! Thank you. Yes, I'm okay ;-) –  amWhy Aug 24 '13 at 12:00