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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

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  • $\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$ Jul 6 at 19:08
  • $\begingroup$ Try noting that $\frac{e^x - 1}{x} = \frac{e^x - e^0}{x-0}$ and relate this to the derivative of $f(x) =e^x$ at a certain point. $\endgroup$
    – angryavian
    Jul 6 at 19:13
  • $\begingroup$ Thanks angryavian $\endgroup$ Jul 6 at 19:40

2 Answers 2

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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.

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  • $\begingroup$ You are faster than me. +1) $\endgroup$
    – xpaul
    Jul 6 at 19:35
  • $\begingroup$ Thanks , I was stucked only with what kind of function should I take , do you have any idea how to take functions while proving such inequality wrt LMVT , Increasing/Decreasing function $\endgroup$ Jul 6 at 20:53
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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.

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  • $\begingroup$ Hi , thanks for the answer, I just have one question regarding your answer you said " one has that f(x) is well-defined in [0,∞)." ... My question is why did you check the limit from right hand side and concluded the function is well defined? .. is it because of singularity ( 1/x tends to infinity as x → 0 ) that the function may not be well defined in neighbourhood of 0? $\endgroup$ Jul 6 at 19:36
  • $\begingroup$ This is because $f(0)$ is not defined. $\endgroup$
    – xpaul
    Jul 6 at 19:42
  • $\begingroup$ If f(0) is not defined, then how f(x) - f(0) = f'(c) (x - 0) would be defined? Because the LHS isn't defined $\endgroup$ Jul 6 at 19:46
  • $\begingroup$ So we have to use the right limit to define $f(0)$. $\endgroup$
    – xpaul
    Jul 6 at 19:48
  • $\begingroup$ I see , thanks sir $\endgroup$ Jul 6 at 19:49

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