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I have to find

$\lim_{n\to +\infty}n\cdot(\sqrt5-\sqrt{5-\frac2n})$

without L'Hospital rule.

I assumed it would be $+\infty\cdot0$ but the book says that it's $\frac{\sqrt5}{5}$ and I have no idea how to reach that result.

Can someone please help me out? Thanks in advance.

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  • $\begingroup$ "I assumed it would be $+\infty\cdot0$" isn't an answer. $+\infty\cdot0$ doesn't mean anything. If you find that a limit "should" tend to something like that, all you've actually found is that you need to dig deeper to find the real answer. $\endgroup$ – Jack M Mar 10 '14 at 9:05
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Try multiplying by the fraction $$\frac{\sqrt{5} + \sqrt{5-\frac2n}}{\sqrt{5} + \sqrt{5-\frac2n}}$$

Then, after canceling, you should obtain an expression where you can let $n$ go to infinity and see what happens.

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This limit can be tranformed to a defintion of the derivative of $2\sqrt{x}$ at $x=5$. $$ \lim_{n\to +\infty}n\cdot\left(\sqrt5-\sqrt{5-\frac2n}\right)=\lim_{n\to +\infty}2\cdot\frac{\sqrt{5-\frac2n}-\sqrt5}{-\frac{2}{n}}=2\lim_{h\to 0}\frac{\sqrt{5+h}-\sqrt5}{h}=2 (\sqrt{x})'_{x=5}\\=2\cdot\frac{1}{2}\cdot \frac{1}{\sqrt{5}}=\frac{1}{\sqrt{5}}. $$

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