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I will appreciate any help with those questions: Prove Using Lagrange's Mean value and Cauchy

$$x+\frac{x^3}{3}>\tan x,\ 0<x<\frac{\pi}{4}$$ I used Lagrange - $f(x)=\tan x$, so there exists $0<c<x$ such that $\displaystyle f'(c)=\frac{\tan(x)}{x}=\frac{1}{\cos^2(c)}$. Next, I was trying to prove that $\displaystyle \frac{1}{\cos^2(c)}<1+\frac{x^2}{3}$, which is usually much more simple but the fact that it has another $x$ in it got me trouble.

Prove: if $f(x)$ is differentiable at $(a,\infty)$ and $\displaystyle \lim_{x\to\infty} f'(x)=0$, then $\displaystyle \lim_{x\to\infty}[f(2x)-f(x)]=0$

By Lagrange, there exists $x<c<2x$ such that $\displaystyle f'(c)=\frac{f(2x)-f(x)}{2x-x}$, hence $\displaystyle \lim_{x\to\infty}\frac{f(2x)-f(x)}{x}=0$, but need to prove without the $x$. I could not find a way to prove it.

Thank you very much!

Ron
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  • Please show us how far you've gotten on your own? –  Feb 20 '16 at 15:03
  • Post edited with more info. Will appreciate any help even on some of them! thanks – Ron Feb 20 '16 at 18:39
  • Nobody has proven those yet? –  Feb 21 '16 at 20:51
  • I easily proved something from the first query, but the $>$ is wrong. In fact, it is: $x+\frac{x^3}{3}<\tan x,\ 0<x<\frac{\pi}{4}$. I have no idea about the second one, though. –  Feb 24 '16 at 20:58

2 Answers2

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  1. I will begin with the second question. It is wrong. Here is a counterexample: $f(x)=\ln x, \forall x\in (0,+\infty )$, then $f$ is differentiable at $(0,+\infty )$ with $f'(x)=\frac{1}{x}, \forall x\in (0,+\infty )$, so that $\lim_\limits{x\to +\infty}{f'(x)}=0$, but $\lim_\limits{x\to +\infty}{\left(f(2x)-f(x)\right)}=\lim_\limits{x\to +\infty}{(\ln (2x) - \ln x)}=\ln2\neq 0$.

  2. For the first question, it is also wrong. We will prove that $x+\frac{x^3}{3}<\tan x,\forall \ 0<x<\frac{\pi}{4}$. Defining $f(x)=x+\frac{x^3}{3}-\tan x,\forall \ 0\leq x< \frac{\pi}{4}$, we can easily see that:

    $$f'(x)=1+x^2-1-\tan^2 x=x^2-tan^2 x=(x+\tan x)(x-\tan x)$$

    Now, let's find the sign of $f'(x)$. We have that $0<x<\frac{\pi}{4} \Rightarrow 0<\tan x<1$, because $\tan x$ is strictly increasing on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, hence $x+\tan x>0, \forall \ 0< x< \frac{\pi}{4}$.

    Defining $g(x)=x-\tan x, \forall \ 0\leq x< \frac{\pi}{4}$, we have that $g'(x)=1-1-tan^2 x=-tan^2x< 0, \forall \ 0< x< \frac{\pi}{4}$, so that $g$ is strictly decreasing on $\left[0, \frac{\pi}{4}\right)$, thus $x>0\Rightarrow g(x)<g(0)=0$.

    Finally, we can now see that $f'(x)<0, \forall \ 0< x< \frac{\pi}{4}$, therefore $f$ is strictly decreasing on $\left[0, \frac{\pi}{4}\right)$, thus $x>0\Rightarrow f(x)<f(0)=0\Rightarrow x+\frac{x^3}{3}<\tan x,\forall \ 0<x<\frac{\pi}{4}$.

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In this answer, it is shown that for $0\le x\le\frac\pi2$, we have $\tan(x)\ge x$. Therefore, $$ \sec^2(x)=1+\tan^2(x)\ge1+x^2 $$ Thus, $$ \begin{align} \tan(x) &=\int_0^x\sec^2(t)\,\mathrm{d}t\\ &\ge x+\frac{x^3}3 \end{align} $$

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