Using LMVT ( Lagrange's Mean Value Theorem) prove the following inequality

Question

I want to prove the following inequality for $x > 0$ $$0 < \frac{1}{x} \log \frac{e^x - 1}{x} < 1$$

1. I think I can do this problem, but I just don't know what should be the function that is to be taken

2. Also , in general is there any trick / way to know what type of function is to be taken to apply lmvt over it ( let's say for proving some other inequality , is there a way to predict/construct a function?)

Note : This problem is an exercise problem from the book "Elements of Real Analysis by Shanti Narayan / MD raisinghania "

Edit : I've managed to do 1) , but I'm still stuck with the 2) query

Answer 1

For $x>0$, the inequality is nothing but proving $1+x < e^x < 1+xe^x$

Let $f(t)=e^t$, apply LMVT for $t\in (0,x).$

We get $\frac{e^x-1}{x}=f'(c)=e^c, c\in (0,x).$

$c\in(0,x) \implies 1<e^c <e^x$

Finally, we get

$1<\frac{e^x-1}{x}<e^x, x>0$ which proves the result.

Answer 2

Let $$ f(x)=\ln\bigg(\frac{e^x-1}{x}\bigg). $$ Since $$ \lim_{x\to0^+}f(x)=\ln\bigg(\lim_{x\to0^+}\frac{e^x-1}{x}\bigg)=0 $$ one has that $f(x)$ is well-defined in $[0,\infty)$. By Lagrange's MVT, one has that, for some $c\in(0,\infty)$ \begin{eqnarray}f(x)-f(0)&=&f'(c)(x-0)\\ &=&x\bigg(\frac{e^c}{e^c-1}-\frac{1}{c}\bigg)\\ &=&x\bigg(\frac{c-1}{c}+\frac{1}{e^c-1}\bigg)\\ &<&x \bigg(\frac{c-1}{c}+\frac{1}{c}\bigg)\\ &=&x \end{eqnarray} and hence the inequality holds. Here $$ e^x>1+x, x>0 $$ is used.