Prove that $0<\dfrac{1}{x}\ln(\dfrac{e^x-1}{x})<1$ for $x>0$ I have the following question before me:
Prove that $0<\dfrac{1}{x}\ln(\dfrac{e^x-1}{x})<1$ for $x>0$ using mean value theorem.
I took $f(x)=\ln(\dfrac{e^x-1}{x})$ and applied LMVT on $f(x)$ on the interval $[0,x]$.
This means that
there exists $c$ in interval $(0,x)$ such that
$\dfrac{f(x)-f(0)}{x-0}=f'(c)$
which further gives
$\dfrac{\ln\dfrac{e^x-1}{x}}{x}= f'(c)$
To prove the desired result, I now need to prove that $0<f'(c)<1$ by proving that $0<f'(x)<1$ for $x>0$.
For this I considered $f'(x)=g(x)=\dfrac{e^x}{e^x-1}-\dfrac{1}{x}$
But I could not really come up with a concrete proof for this.
You are requested to help me with this. Any help will be greatly appreciated.
 A: Upon simplification we have to prove that $$1+x< e^x<\frac{1}{1-x}, \forall~ x\in (0,1)$$
Let $f(x)=e^x-1-x \implies f'(x)=e^x-1>0, \forall ~ x\in (0,1).$
This means $f(x)$ is an increasing function, so $f(x)>f(0) \implies e^x>1+x.$
Next, let $g(x)=(1-x)e^x-1\implies g'(x)=-e^x+(1-x)e^x=-xe^x <0$, so $g(x)$ is a decreasing function which implies $g(x)<g(0) \implies (1-x)e^x <1 \implies e^x <\frac{1}{1-x}$.
A: I think you're supposed to apply the MVT to $e^x$, using that $e^x - 1 = e^x - e^0 = x e^a$ for some $0 < a < x$. Then ${e^x - 1 \over x} = e^a$. Try taking it from there.
A: Equivalently, one has to prove that $$1+x<e^x<1+xe^x,\quad x\in (0,1). \tag{1}$$
Apply LMVT to $f(t)=e^t$, $\;t\in (0,x)$
$$\frac{e^x-1}{x-0}=f'(c)=e^c,\quad 0 <c<x. \tag{2}$$
Next, note that  $0<c<1 \implies 1 <e^c<e^x$, then from $(2)$, we can write
$$1<\frac{e^x-1}{x-0}<e^x,\quad x\in(0,1).$$
LHS gives both the results in $(1)$.
A: Note that for $x>0$,
$$
1 + x < 1 + x + \frac{{x^2 }}{{2!}} + \frac{{x^3 }}{{3!}} +  \ldots  < 1 + x + x^2  + \frac{{x^3 }}{{2!}} +  \ldots ,
$$
that is $1+x<e^x<1+xe^x$ for all $x>0$. This is equivalent to the desired inequality.
