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I have read Why $x<\tan{x}$ while $0<x<\frac{\pi}{2}$?

If I want to get $\tan{x}\gt x$ instead of weaker inequality $\tan{x} \ge x$. Do I need only to show that $\tan{x} \gt x$ when $x\to 0$? Because from @David Mitra 's picture, it is obvious to see $\tan{x}\gt x$ when $x$ is not near $0$.

Since $$\lim_{x\to 0}\frac{\tan{x}}{x}=1 \text{}$$ ,for $$\varepsilon=\frac{1}{n} \;\;\text{ where }n\in\Bbb N$$, we can find $\delta \gt 0 $ s.t. $$\forall x \in (0,0+\delta)$$, we have $$\frac{\tan{x}}{x}\gt1-\frac{1}{n}$$

Let $n\to \infty$, we get $$\frac{\tan{x}}{x}\gt1$$ So $$\tan{x}\gt x \text{ when }x\in(0,\frac{\pi}{2})$$

Or someone has more analytical proof instead of geometric proof for $x\in(0,\frac{\pi}{2})$, since I only prove the case $x$ is near the origin. Thanks for helping.

Jean Marie
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Steven Lu
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2 Answers2

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Set $$f(x)=\tan(x)-x.$$ One can prove that $f'(x)>0$ on $(0,\frac{\pi}{2})$, and thus $f$ is strictly increasing on $(0,\frac{\pi}{2})$. Therefore $$f(x)> f(0)=0,$$ for all $x\in (0,\frac{\pi}{2}).$

Surb
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It is a direct consequence of the Mean value theorem:

For any $x\in \bigl(0,\frac\pi2\bigr)$, we have $$\frac{\tan x}x=(\tan)'(\xi)\quad(0<\xi<x)\quad =\frac1{\cos^2\xi},$$ and on the interval $\bigl(0,\frac\pi2\bigr)$, $\;0<\cos\xi<1$, so $$\frac{\tan x}x=\frac1{\cos^2\xi}>1.$$

Bernard
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