Show by $\varepsilon-\delta$ argument that $\displaystyle\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$.

Question

How to show that $\displaystyle\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$ by $\varepsilon-\delta$ argument?

In particular, students who prove this have not learned the Taylor theorem and series.

I think I need to use the Mean Value Theorem to create a suitable inequality, but I don't have any idea.

Can someone help me? Thank you in advance.

Answer 1

Alright so you mentioned in the comments that you're using the definition

$$\ln x=\int_1^x\frac{\mathrm{d}t}{t}.$$

Let's start by establishing some inequalities. Notice first that $\frac{1}{t}\leq1$ for all $t\in[1,\infty)$, and so

$$\ln x=\int_1^x\frac{\mathrm{d}t}{t}\leq\int_1^x\mathrm{d}t=x-1$$

for all $x\geq 1$. Secondly, if we substitute $u=\frac{1}{t}$, $\mathrm{d}u=-\frac{\mathrm{d}t}{t^2}$ in the integral, then

$$\ln x=-\int_1^{\frac{1}{x}}\frac{\mathrm{d}u}{u}=\int_{\frac{1}{x}}^1\frac{\mathrm{d}u}{u}\geq\int_{\frac{1}{x}}^1\mathrm{d}u=1-\frac{1}{x}$$

for all $x\geq 1$, as $\frac{1}{u}\geq 1$ for all $u\in(0,1]$. By similar arguments you can establish that these inequalities hold for all $x>0$. Now with these inequalities at hand we get that

$$\frac{1}{1+x}\leq\frac{\ln(1+x)}{x}\leq1.$$

From here just use the squeeze theorem (or do a similar argument to the proof of it) to get the result.

Answer 2

First take $x>0$. $|\frac {\ln (1+x)} x -1|=|\frac 1 x \int_1^{1+x} (\frac 1 t -1)dt|\leq \frac 1 {x}\int_1^{1+x} \frac {|1-t|}{t}dt$. Note that $|1-t| \leq x<\epsilon$ if $0<x<\epsilon$. Also, $\frac 1 {x}\int_1^{1+x} \frac 1 {t}dt\leq \frac 1 {x}\int_1^{1+x} dt=1$. I will leave the case $x<0$ to you.