$\lim\limits_{N \to +\infty} \sqrt{N+1} \log\left (1+\frac{x}{N+1}\right)$ 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?
 A: 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
$0 < c < 1$.
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
$.
A: 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. 
A: If we are allowed to use L' Hospital:
$$\lim_{n\to \infty}\frac{\ln(1+\frac{x}{n+1})}{\frac{1}{\sqrt{n+1}}} = \lim_{n\to \infty}\frac{2(n+1)^{5/2}}{(n+1+x)(n+1)^2} \to 0$$
A: 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
$$\lim_{z \to 0} x \left(\frac{\log\left(1 + xz^2\right)}{xz^2}\right) = x \cdot 1 = x.$$
A: You can write:
$$\lim_{n\to\infty}(\sqrt{n+1}\ln(1+\frac x{n+1})) = \lim_{n\to\infty}\frac{(n+1)\ln(1+\frac x{n+1})}{\sqrt{n+1}} = \lim_{n\to\infty}\frac{\ln(1+\frac x{n+1})^{n+1}}{\sqrt{n+1}}. $$
Can you finish from here?
A: hint: use taylor expansion for the Log term
