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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.

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Try

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

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

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