Let's pose $z = \frac{1}{\sqrt{N+1}}$. Then, when $N$ goes to $+\infty$, then $z$ goes to $0$. You can rewrite your limit as follows:
$$\lim_{z \to 0} \frac{1}{z} \log\left(1 + xz^2\right).$$
It is well known that:
$$\lim_{a \to 0} \frac{\log\left(1 + a\right)}{a} = 1.$$
Starting from this, we can rewrite the original limit as follows:
$$\lim_{z \to 0} xz \left(\frac{\log\left(1 + xz^2\right)}{xz^2}\right) = 0 \cdot 1 = 0.$$
The idea here is that "$a = xz^2$", since both goes to $0$...
If the limit was
$$\lim\limits_{N \to +\infty} (N+1) \log \left(1+\frac{x}{N+1}\right),$$
then, passing to $z$, we get:
$$\lim_{z \to 0} \frac{1}{z^2} \log\left(1 + xz^2\right),$$
or equivalently
$$\lim_{z \to 0} x \left(\frac{\log\left(1 + xz^2\right)}{xz^2}\right) = x \cdot 1 = x.$$