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?

  • 3
    $\begingroup$ $$\log (1 + t) = 1 + t + O(t^2)$$ $\endgroup$ – user296602 Oct 21 '17 at 19:25
  • 1
    $\begingroup$ Letting $M=N+1$ will at least reduce the noise in your equation :) $\endgroup$ – Thomas Andrews Oct 21 '17 at 19:43
  • 2
    $\begingroup$ $$\log (1 + t) = {\color{red}0} + t + O(t^2)$$ $\endgroup$ – the_candyman Oct 22 '17 at 8:31

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 $.


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.


If we are allowed to use L' Hospital:

$$\lim_{n\to \infty}\frac{\ln(1+\frac{x}{n+1})}{\frac{1}{\sqrt{n+1}}} = \lim_{n\to \infty}\frac{2(n+1)^{5/2}}{(n+1+x)(n+1)^2} \to 0$$


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

$$\lim_{z \to 0} x \left(\frac{\log\left(1 + xz^2\right)}{xz^2}\right) = x \cdot 1 = x.$$

  • $\begingroup$ Good rephrasing. $\endgroup$ – marty cohen Oct 22 '17 at 19:05
  • $\begingroup$ @martycohen Good answer (+1) $\endgroup$ – the_candyman Oct 22 '17 at 19:59

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?


hint: use taylor expansion for the Log term

  • $\begingroup$ Why the down votes to this answer? The OP asked for clue; this is a fine clue, without huffing and puffing. $\endgroup$ – kimchi lover Oct 21 '17 at 19:54

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