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

Question

In proof of $\displaystyle\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1$ is assumed that $\sin{x}\leq{x}\leq\tan{x}$ while $0<x<\frac{\pi}{2}$. First comparison is clear, arc length must be greater than sine value, but how about $x\leq\tan{x}$, why tangent is longer than arc?

Answer 1

It's fine if the comparisons of those lengths is intuitively clear to you, but if you want to be rigorous, it's easier to compare nested areas than it is to compare curve lengths.

Working off of David Mitra's picture (see his answer), the area of the triangle spanned by the lines $\cos(t)$ and $\sin(t)$ is $\frac{1}{2} \cos(t)\sin(t)$. That area rests inside the sector of angle $t$, which has area $\frac{t}{2}$ (the proportion $\frac{t}{2\pi}$ of the entire circle's area $\pi$). And in turn, the area of sector is inside the triangle spanned by the horizontal radius of the unit circle and $\tan(t)$, which has area $\frac{1}{2} \tan(t)$.

Thus we have the inequality $$\frac{1}{2} \cos(t)\sin(t) \le \frac{t}{2} \le \frac{1}{2} \tan(t)$$ Or $$\cos(t)\sin(t) \le t \le \tan(t)$$ canceling the 2's. You should be able to adjust your proof of $\displaystyle\lim_{t \to 0} \frac{\sin(t)}{t} = 1$ to utilize this slightly weaker inequality.

Answer 2

Since the first comparison is clear to you....If you draw the unit circle and recognize the side of tangent, then the second one is also clear to you.

Answer 3

My approach is probably different from what what you want to hear:

The derivative of $\tan x$ is 1 when $x = 0$ and is increasing, it can be shown easily that $\tan x$ is (strictly) convex for $x \in (0, \pi/2)$. And since $y=x$ is its tangent at the point $[0,0]$, the inequality $\tan x > x$ has to hold. I don't know if you discussed tangents and convex/concave functions, but it is one of the basic properties.

Hope I helped at least a bit. Cheers.

Answer 4

$\ \ \ \bullet$ Using similar triangles: $$\color{darkgreen}{\tan t}={\color{maroon}{\sin t}\over\color{darkblue}{\cos t}} ={ {\text{length}( \color{darkgreen} {\overline{{IZ}})} }\over 1 }\quad \Longrightarrow \quad\color{darkgreen}{\tan t}=\text{length}(\color{darkgreen}{\overline{IZ}})$$

$\ \ \ \bullet$ $t$ is the length of the arc $\color{orange}{IQ}$.

$\ \ \ \bullet$ Area of the circular sector $O\color{orange}{IQ}={t\over 2\pi}\cdot \pi\cdot 1^2={t\over2}$.

$\ \ \ \bullet$ Area of $\triangle OIZ={1\over2}\cdot1\cdot\color{darkgreen}{\tan t}$.

So $$\color{maroon}{\sin t}\lt t\lt\color{darkgreen}{\tan t}$$ for $0< t<\pi/2$.

Answer 5

You could also use calculus.
Let $f(x)=x-\tan(x)$, $0\lt x \lt \frac{\pi}{2}$. Then $$f'(x)=1-\sec^2(x)=-\tan^2(x) \leqslant 0~,~~~~~0\lt x\lt \frac{\pi}{2},$$ So $f(x)$ decreases on $0\lt x\lt \frac{\pi}{2}$. Thus we have $f(0)\gt f(x)$ or $f(x)\lt 0$.