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

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?

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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.

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surprisingly that was the best explanation for me, thanks! –  haemhweg Jan 14 '12 at 14:43

$\ \ \ \bullet$ 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 $0< t<\pi/2$.

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The last two bullet points were added after @Dustan Levenstein 's post reminded me that the fact that the length of the circular arc $IQ$ is less than $\tan t$ required some explanation. (It should be clear that $\sin t$ is less than the length of the line segment $\overline{QI}$, which in turn is less than the length of the arc $QI$.) –  David Mitra Jan 14 '12 at 14:54
+1 for the use of matching colors in the picture and equations. Which program did you use to produce the picture? –  Kavka Jan 14 '12 at 15:11
@Kavka A pain to use really, but I like it: jsxgraph.org –  David Mitra Jan 14 '12 at 15:14

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.

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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.

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OP wanted to prove $\displaystyle\lim_{x\to 0} \frac{\sin(x)}{x} = 1$. I think the trigonometric derivatives are still a little ways down the road. –  Dustan Levenstein Jan 14 '12 at 14:37

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 $f(x)\lt 0$.

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I already pointed this out on Ondrej's answer. As long as we're willing to use calculus, why not just take the derivative of $\sin(x)$ and plug in $x=0$? –  Dustan Levenstein Jan 14 '12 at 14:52