Prove that $\lim_{x \rightarrow +\infty} \frac{\text{log}(x+1)}{x} = 0$ without de l'Hopital rule

Question

$\lim_{x \rightarrow +\infty} \frac{\text{log}(x+1)}{x} = 0$

I know that $\text{log}(x+1)$ goes to infinity more slowly than $x$, but I'm not able to prove the limit above without using de l'Hopital rule.

Ideally, the proof should be algebraic and not with a graphic showing that $\text{log}(x+1)$ goes to infinity more slowly than $x$.

Answer 1

Since$$\lim_{x\to\infty}\frac{\frac{\log(x+1)}x}{\frac{\log(x)}x}=\lim_{x\to\infty}\frac{\log(x)+\log\left(1+\frac1x\right)}{\log(x)}=1,$$it is enough to prove that$$\lim_{x\to\infty}\frac{\log(x)}x=0.$$Doing $x=e^y$, this becomes$$\lim_{y\to\infty}\frac y{e^y}=0.$$But, if $y>0$,$$e^y=1+y+\frac{y^2}2+\cdots>\frac{y^2}2,$$and therefore$$0<\frac y{e^y}<\frac y{y^2/2}=\frac2y\to0.$$

Answer 2

Note that for $x\gt 0$, $\log(1+x)=2\log \sqrt{x+1}\le 2\sqrt {x+1}$

On dividing the inequality by $x$, we get for large $x$

$0\le \frac{\log(1+x)}x\le2\sqrt{\frac 1x+\frac 1{x^2}}$

The result follows by squeeze principal.

Note: The inequality used in first line follows from the famous inequality for all $t\gt 0$, $\log t\le t$.

A short proof (using application of derivatives) for the inequality could go along these lines:

Let $g(x)=\log t-t$. It would suffice to show that $g(x)\le 0$ for all $x\gt 0$. Clearly for $0\lt t\le 1$ the inequality holds as $\log t$ is non positive on $(0,1]$. So we can concentrate on $t\gt 1$ and noting that derivative of $g$ that is $g’(t)\lt 0$ for all $t\gt 1$, it follows that $g$ is decreasing on $[1,\infty)$ so $g(t)\lt g(1)=0$ for all $t\gt 1$ and hence the inequality follows.

Answer 3

One more method trying to use only elementary facts.

Suppose, for $a>1$, we have $$\lim\limits_{n\to\infty}\frac{\log_a n}{n}=0\quad(1)$$ and let's consider any $x_n\to +\infty$. As we have $\left\lfloor x_n \right\rfloor \leqslant x_n \leqslant \left\lfloor x_n \right\rfloor+1$, then, for appropriate $n$, holds

$$ 0\leqslant \frac{\log_a x_n}{x_n} \leqslant \frac{\log_a ( \left\lfloor x_n \right\rfloor+1)}{\left\lfloor x_n \right\rfloor}=\frac{\log_a ( \left\lfloor x_n \right\rfloor+1)}{\left\lfloor x_n \right\rfloor+1}\cdot\frac{ \left\lfloor x_n \right\rfloor+1}{\left\lfloor x_n \right\rfloor}$$ where in right hand we have subsequences of $(1)$ and $\frac{n+1}{n}\to 1$, so right side tends to zero.

Now for $(1)$ I assume we know $\lim\limits_{n\to\infty}\frac{n}{b^n}=0$, for $b>1$, based on binomial theorem. So, for appropriate $n$, we have $\frac{1}{b^n} < \frac{n}{b^n}<1$. Let's take $b=a^\varepsilon$, where $a>1$ and $\varepsilon > 0$. Then we obtain $\frac{1}{a^{n\varepsilon}} < \frac{n}{a^{n\varepsilon}} < 1$, or, same is $1<n<a^{n\varepsilon}$. From last we obtain $$0< \log_a n < n\varepsilon$$ and at last $$0< \frac{\log_a n}{n} < \varepsilon$$ which gives $(1)$.

Answer 4

Here is an answer that only uses basic properties of $\log(n)$ and Bernoulli's Inequality.

For $n\ge3$, $$ \begin{align} (n+1)^{n+1} &=n^{n+1}\left(1+\frac1n\right)^{n+1}\tag{1a}\\ &\le4n^{n+1}\tag{1b}\\ (n+1)^n &\le\frac4{n+1}n^{n+1}\tag{1c}\\ &\le n^{n+1}\tag{1d}\\ \frac{\log(n+1)}{n+1} &\le\frac{\log(n)}n\tag{1e} \end{align} $$ Explanation:
$\text{(1a)}$: distributive property
$\text{(1b)}$: as shown in this answer, using only Bernoulli's Inequality,
$\phantom{\text{(1b):}}$ $\left(1+\frac1n\right)^{n+1}$ is decreasing; thus, for $n\ge1$, we have $\left(1+\frac1n\right)^{n+1}\le4$
$\text{(1c)}$: divide by $n+1$
$\text{(1d)}$: $n\ge3$
$\text{(1e)}$: take logs and divide by $n(n+1)$

$(1)$ shows that $\frac{\log(n)}{n}$ is decreasing for $n\ge3$. Since $\frac{\log(n)}{n}$ is bounded below by $0$, $L=\lim\limits_{n\to\infty}\frac{\log(n)}{n}$ exists.

Furthermore, $$ \begin{align} L &=\lim_{n\to\infty}\frac{\log(2n)}{2n}\tag{2a}\\ &=\lim_{n\to\infty}\frac{\log(2)}{2n}+\frac12\lim_{n\to\infty}\frac{\log(n)}n\tag{2b}\\ &=0+\frac12L\tag{2c} \end{align} $$ Explanation:
$\text{(2a)}$: definition
$\text{(2b)}$: $\log(2n)=\log(2)+\log(n)$
$\text{(2c)}$: take the limits

$(2)$ says that $\lim\limits_{n\to\infty}\frac{\log(n)}{n}=L=0$. Thus, $$ \begin{align} \lim_{n\to\infty}\frac{\log(n+1)}n &=\lim_{n\to\infty}\frac{\log(n+1)}{n+1}\lim_{n\to\infty}\frac{n+1}{n}\tag{3a}\\[6pt] &=0\cdot1\tag{3b}\\[9pt] &=0\tag{3c} \end{align} $$

Answer 5

Reversing the order of operations we have:

$\text{log}\lim_{x \rightarrow +\infty}(x+1)^\frac{1}{x}=$,

$log(1)=0$.