# Limit $\lim\limits_{n \to \infty} n \cdot\ln(\sqrt{n^2+2n+5}-n)$

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.

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

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 $$\lim_{a \to \infty} a \cdot \frac{2}{a} = 2$$
• @Assaultuous2 Sorry, may be its a stupid question, but can you please explain me, why $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? – Ivan Gandacov Oct 14 '14 at 17:42
• It means that for $x \to 0$ $\log (1+x)$ grows at roughly the rate of $x$. You can get it using Taylor series. – Alex Oct 14 '14 at 18:21
• More precisely: $\lim_{x\to0}\frac{\ln(1+x)}x=1$ (also written as $\ln(1+x)\sim x$ as $x\to0$). – Akiva Weinberger Oct 14 '14 at 18:31