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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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6 Answers 6

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$\ \ \ \bullet$ 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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  • $\begingroup$ 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 $QI$.) $\endgroup$ Jan 14, 2012 at 14:54
  • $\begingroup$ +1 for the use of matching colors in the picture and equations. Which program did you use to produce the picture? $\endgroup$
    – Kavka
    Jan 14, 2012 at 15:11
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    $\begingroup$ @Kavka A pain to use really, but I like it: jsxgraph.org $\endgroup$ Jan 14, 2012 at 15:14
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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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  • $\begingroup$ surprisingly that was the best explanation for me, thanks! $\endgroup$
    – nakajuice
    Jan 14, 2012 at 14:43
  • $\begingroup$ I'm not much of a math guy, but doesn't this require $t$ to be within $[0, \frac{\pi}{2}]$? How then does it prove the limit for all $t$? $\endgroup$ Oct 7, 2021 at 3:12
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    $\begingroup$ @marsnebulasoup You're right, this proves the limit from the rhs, and you can use negative angle identities to get the limit from the lhs. Note that you only need to know about values near $0$, so $t>\frac{\pi}{2}$ is irrelevant. $\endgroup$ Oct 11, 2021 at 15:26
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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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    $\begingroup$ 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 $x=0$? $\endgroup$ Jan 14, 2012 at 14:52
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    $\begingroup$ Any and all calculus regarding trigonometric functions has to use the result $\lim_{x \to 0} \frac{\sin{x}}{x} = 1$. So this method just shows consistency and is circular reasoning $\endgroup$ Sep 10, 2019 at 8:51
  • $\begingroup$ isn't it circular reasoning as derivatives are itself derived from limits $\endgroup$ Nov 30, 2022 at 3:42
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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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    $\begingroup$ 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. $\endgroup$ Jan 14, 2012 at 14:37
  • $\begingroup$ @ondrej,your approach is correct if you assume derivative of tanx=1 when x=0,as it was itself derived from these limits the OP is asking for $\endgroup$ Nov 30, 2022 at 3:43
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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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The inequality $$\sin x < x < \tan x $$ should be clear from the picture below

Bieberbach

It is from the book Differentialrechnung I by Ludwig Bieberbach.

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