# Calculating a limit with Floor Function

Can someone help me understand how to calculate the limit: $\lim_{n \to \infty} n [\sqrt{n+4}-\sqrt{n} ]$ ?

How can I det rid the floor function ?! (Multiplying by $\sqrt{n+4}+\sqrt{n}$ gives me nothing.

Can you help me?

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You see, as n goes to infinity, the difference between the $\sqrt{n+4}-\sqrt{n}$ becomes less than 1, so their floor function is $0$ – ciceksiz kakarot Feb 10 '13 at 20:32

Hint: for large $n$ $\sqrt{n+4}-\sqrt{n}=\frac{(\sqrt{n+4}-\sqrt{n})(\sqrt{n+4}+\sqrt{n})}{\sqrt{n+4}+\sqrt{n}}=\frac{4}{\sqrt{n+4}+\sqrt{n}}$ is very low (less than 1). Thus: $$\left \lfloor\sqrt{n+4}-\sqrt{n}\right \rfloor=0$$ $$n\left \lfloor\sqrt{n+4}-\sqrt{n}\right \rfloor=0$$ $$\lim_{n\rightarrow\infty}n\left \lfloor\sqrt{n+4}-\sqrt{n}\right \rfloor=0$$

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Actually, not for "large" - already $n\geq 3$ is good enough ;-) – Johnny Westerling Feb 10 '13 at 20:34
@Johnny Westerling Yes! I often say large( or sufficently large) instead of saying there exists $n$ – Amr Feb 10 '13 at 20:36

$$\sqrt{n+4}-\sqrt n=\frac{4}{\sqrt{n+4}+\sqrt n}\xrightarrow [n\to\infty]{}0\Longrightarrow$$

for $\,n\,$ big enough (say, $\,n>7\,$) we get $\,\sqrt{n+4}-\sqrt n<1\Longrightarrow [\sqrt{n+4}-\sqrt n]=0\,$ , and thus the limit is zero.

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Use estimate $$0< \sqrt{n+4}-\sqrt{n} =\dfrac{n+4-n}{\sqrt{n+4}+\sqrt{n}}=\dfrac{4}{\sqrt{n+4}+\sqrt{n}}\leqslant \dfrac{4}{2\sqrt{n}}= \dfrac{2}{\sqrt{n}},$$ so, for $n\geqslant{5}$ $$[\sqrt{n+4}-\sqrt{n}]=0.$$

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