$\lim\limits_{N \to +\infty} \sqrt{N+1} \log\left (1+\frac{x}{N+1}\right)$ I have to compute the limit

$$
\lim\limits_{N \to +\infty} \sqrt{N+1} \log \left(1+\frac{x}{N+1}\right)
$$
where $x \ge 0$ is fixed. I tried to see the previous as

$$
\log \lim\limits_{N \to +\infty} \left(1+\frac{x}{N+1}\right)^{\sqrt{N+1}}
$$
and to change variable, but it doesn't work. Intuitively, this limit is 0, but I have no clue on how to solve it. Can you help me?
The "+1" doesn't matter.

For large $N$ and fixed $x$

$\sqrt{N} \log (1+\frac{x}{N})
=\sqrt{N} (\frac{x}{N}+O((\frac{x}{N})^2))
=\frac{x}{\sqrt{N}}+O(\frac{1}{N^{3/2}})
\to 0
$.

This holds with
$N^{c}$
(instead of $\sqrt{N}$)
for all
$0 &lt; c &lt; 1$.

If
$c=1$,
we have
$N \log (1+\frac{x}{N})
=N (\frac{x}{N}+O((\frac{x}{N}))^2)
=x+O(\frac{1}{N})
\to x
$.
hint: use taylor expansion for the Log term