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How should this limit be solved ? $$\lim_{n \to \infty} n \cdot \ln(\sqrt{n^2+2n+5}-n)$$

I've tried to multiply and at the same time divide $\sqrt{n^2+2n+5}-n$ by $\sqrt{n^2+2n+5}+n$, and then make $n$ as the power of $\frac {2n+5}{\sqrt{n^2+2n+5}+n}$. But I got stuck. I dont think it was the best idea.

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    $\begingroup$ Do you know L'Hôpital's rule? $\endgroup$ Commented Oct 14, 2014 at 17:07
  • $\begingroup$ @columbus8myhw Yes its is used in cases of nedetermination as 0/0 or $\frac{\infty}{\infty}$ $\endgroup$ Commented Oct 14, 2014 at 17:13
  • $\begingroup$ You have $\infty\times0$, which can be easily manipulated into $\frac00$. $\endgroup$ Commented Oct 14, 2014 at 17:23

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Notice that $$ \lim_{n \to \infty} n \cdot \ln (\sqrt{n^2+2n+5}-n) = \lim_{n \to \infty} n \cdot \ln (\sqrt{(n+1)^2+4}-n) $$ so we change variable: $a= n+1 \to \infty $ to get $$ \lim_{a \to \infty} (a-1) \cdot \ln (\sqrt{a^2+4}-a+1) $$ But we know that for $a \to \infty $, for a product $ (a-1)$ behaves as $a$ and that $ \sqrt{a^2+4} $ behaves as $ a + \frac{2}{a} $ and so our limit becomes $$ \lim_{a \to \infty} a \cdot \ln (a+\frac{2}{a}-a+1) $$ which simplifies to $$ \lim_{a \to \infty} a \cdot \ln (1+\frac{2}{a}) $$ and since, for small $x$, one has $ \ln(1+x) = x $, the limit becomes $$ \lim_{a \to \infty} a \cdot \frac{2}{a} = 2 $$

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  • $\begingroup$ Furthermore, using a calculator to determine the value of the expression for n=1000 gives a value of 1.996, which shows the expression tending to 2... $\endgroup$ Commented Oct 14, 2014 at 17:42
  • $\begingroup$ @Assaultuous2 Sorry, may be its a stupid question, but can you please explain me, why $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? $\endgroup$ Commented Oct 14, 2014 at 17:42
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    $\begingroup$ The square of a+2/a is a^2+4+4/a^2, which is what is inside your root, just off by a very tiny value of order 1/a^2. $\endgroup$ Commented Oct 14, 2014 at 17:47
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    $\begingroup$ It means that for $x \to 0$ $\log (1+x)$ grows at roughly the rate of $x$. You can get it using Taylor series. $\endgroup$
    – Alex
    Commented Oct 14, 2014 at 18:21
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    $\begingroup$ More precisely: $\lim_{x\to0}\frac{\ln(1+x)}x=1$ (also written as $\ln(1+x)\sim x$ as $x\to0$). $\endgroup$ Commented Oct 14, 2014 at 18:31

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