Show $\tan(x)-x>0$ ,$\forall x \in (0,\pi/2)$

Question

I know the derivative is greater than $0$ for all $x$ in $(0, \pi/2)$, but how to show $\tan(x)-x $ is greater than $0$ as $x$ approaches $0$?

Note: we have not yet learned l'hospitals rule.

Answer 1

Let $f(x) = \tan(x) - x$.

Then $f'(x) = \sec^2(x) - 1 = \frac{1-\cos^2(x)}{cos^2(x)} = \tan^2(x) \geq 0 \quad \forall x \in (0, \frac{\pi}{2})$

Hence $f(x) \geq f(0) = \tan(0) - 0 = 0 \quad \forall x \in (0, \frac{\pi}{2})$

Answer 2

Let $f(x)=\tan x-x$. Then $f'(x)=\sec^2x-1=\tan^2x > 0$ when $x\in (0, \frac{\pi}{2}).$

So $f(x)=\tan x-x\ge 0$, i.e., $\tan x > x$ when $x\in (0, \frac{\pi}{2}).$

Answer 3

Define $f(x) = \tan(x) - x$. First, we will prove that function is increasing on $(0, \pi/2)$. For any $x \in (0, \pi/2)$, we have $0 < \cos(x) < 1$. This means that $\sec^2(x) - 1 > 0$. But, we have $f'(x) = \sec^2(x) - 1$. Hence $f'(x) > 0$ for $x \in (0, \pi/2)$.

Now, let $x_1 \in (0, \pi/2)$ be some arbitrary point. Consider the closed interval $[0, x_1]$. $f(x)$ is continuous over $[0, x_1]$ and differentiable over $(0, x_1)$. Hence by Mean Value theorem, there is some $c \in (0, x_1)$ such that

$$ f'(c) = \frac{f(x_1) - f(0)}{x_1 - 0} = \frac{\tan(x_1) - x_1}{x_1} $$

Since $f$ is an increasing function over $(0, \pi/2)$, we have $f'(c) > 0$. Hence $\tan(x_1) > x_1$.

Since $x_1$ is some arbitrary point in $(0, \pi/2)$, it proves that

$$ \tan(x) - x > 0 \qquad \forall x \in (0, \pi/2) $$

Answer 4

As an alternative, using the same idea used here to prove the standard limit $\frac{\sin x}x$, by a purely geometric approach according to the following figure

we have that $\forall x \in (0,\pi/2)$

• area of yellow wedge $OAB=\frac x2\cdot 1^2=\frac x 2$

is less than

• area of $\triangle OAC=\frac12 \cdot 1 \cdot \tan x= \frac{\tan x }2$

that is $\tan x -x>0$.