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.

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
$$ \lim_{a \to \infty} a \cdot \frac{2}{a} = 2 $$
