$\lim\limits_{N \to +\infty} \sqrt{N+1} \log\left (1+\frac{x}{N+1}\right)$ 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?

This one is easy and an immediate consequence of the fundamental inequality satisfied by $\log$ function: $$\log x\leq x-1,x>0\tag{1}$$ For the current question we have $x\geq 0$ and hence $$0\leq \sqrt{N+1}\log\left(1+\frac{x}{N+1}\right) \leq \frac{x} {\sqrt{N+1}}$$ and the result follows via Squeeze Theorem. 
