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Using the Algebra of Limits, Calculate the following limit L:

$$\lim_{n\to \infty}\left(\sqrt{1+n}-\sqrt{n}\right)\sqrt{n+\frac 12}=L$$

I have tried applying the distributive law and then seperated the limit using the property of subtraction of limits but I am not sure if this is the right way to take on this question. How do I need to start this question off?

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Hint. Try the standard calculation: $$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}.}$$

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\begin{align*} (\sqrt{1+n}-\sqrt n)\sqrt{n+\frac12} &=\frac{(\sqrt{1+n}-\sqrt n)(\sqrt{1+n}+\sqrt n)}{\sqrt{1+n}+\sqrt n}\sqrt{n+\frac12}\\ &=\frac{\sqrt{n+\frac12}}{\sqrt{1+n}+\sqrt n}\\ &=\frac{\sqrt{1+\frac1{2n}}}{\sqrt{\frac1n+1}+1}\\ &\to\frac12 \end{align*} as $n\to\infty$.

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