How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.
This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.
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You should first prove that for $x > 0$ small that $\sin x < x < \tan x$. Then, dividing by $x$ you get $$ { \sin x \over x} < 1 $$ and rearranging $1 < {\tan x \over x} = {\sin x \over x \cos x }$ $$ \cos x < {\sin x \over x}. $$ Taking $x \rightarrow 0^+$ you apply the squeeze theorem. For $x < 0$ and small use that $\sin(-x) = -\sin x$ so that $${\sin(-x) \over -x} = {\sin x \over x}.$$ As far as why the first inequality I said is true, you can do this completely from triangles but I don't know how to draw the pictures here.
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It depends on your definition of the sine function. I would suggest checking out the geometric proof in ProofWiki.
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Usually calculus textbooks do this using geometric arguments followed by squeezing.
Here's an Euler-esque way of looking at it---not a "proof" as that term is usually understood today, but still worth knowing about.
Let $\theta$ be the length of an arc along the circle of unit radius centered at $(0,0)$, from the point $(1,0)$ in a counterclockwise direction to some point $(\cos\theta,\sin\theta)$ on the circle. Then of course $\sin\theta$ is the height of the latter point above the $x$-axis. Now imagine what happens if $\theta$ is an infinitely small positive number. Then the arc is just an infinitely short vertical line, and the height of the endpoint above the $x$-axis is just the length of the arc. I.e. when $\theta$ is an infinitely small number, then $\sin\theta$ is the same as $\theta$. It follows that when $\theta$ is an infinitely small nonzero number, then $\dfrac{\sin\theta}{\theta}=1$.
That is how Euler viewed the matter. See his book on differential calculus.
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Look at this link:
http://fatosmatematicos.blogspot.com/2010/08/provas-sem-palavras-parte-20.html
Here is the picture I copied from that blog:
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Here you may see an elementary approach that starts from a very interesting result, see this problem. All you need is a bit of imagination. When we take $\lim_{n\rightarrow\infty} \frac{n\sin(\frac{\pi}{n})}{1+\sin(\frac{\pi}{n})}$ we may notice that we have infinitely many circles surrounding the unit circle with infinitely small diameters that lastly perfectly approximate the length of the unit circle when having it there infinity times . Therefore when multiplying n by the radius under the limit to infinity we get π. Let's denote $\frac{\pi}{n}$ by x.
$$\lim_{x\rightarrow0}\frac{\pi\sin(x)}{x(1+\sin(x))}={\pi}\Rightarrow\lim_{x\rightarrow0}\frac{\sin(x)}{x(1+\sin(x))}=1\Rightarrow\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$$
The proof is complete.
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This is not a rigorous proof, but is instead an intuitive argument. Consider the graph of the sine function and in particular consider the origin $(0,0)$ and some arbitrary point $(x,\sin(x))$ a little bit to the right of it. Connect a secant line between the two points like so:
Now consider the average rate of change of the sine function on that interval $[0,x]$, or alternatively the slope $m$ of that secant line, namely:
$$m(x)=\frac{\sin(x)-\sin(0)}{x-0}$$
But clearly this is just
$$m(x)=\frac{\sin(x)}{x}$$
But then "clearly" (again, this is just for intuitive purposes), as $x\to{0}$, the slope of this secant line approaches the slope of the tangent line to the graph at $x=0$ (i.e. it approaches whatever $\sin'(0)$ should be). Looking at the graph then, it shouldn't feel too unreasonable that this limiting procedure results in a $45$-degree diagonal tangent, hence a slope of $1$.
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This is a variant of robjohn's answer. The area of the sector $ADE$ is $\frac{1}{2}x\cos^2(x)$; the area of the triangle $ABC$ is $\frac{1}{2}\sin(x)$; and the area of the sector $ABC$ is $\frac{1}{2}x$. By inclusion, we find that for $0<x < \frac{\pi}{2}$, $$ \frac{1}{2}x\cos^2(x) \le \frac{1}{2}\sin(x) \le \frac{1}{2}x \, . $$ If we multiply each term in this inequality by $\frac{2}{x}$, we obtain the necessary bounds to apply the squeeze theorem: $$ \cos^2(x) \le \frac{\sin(x)}{x} \le 1 \, . $$ This inequality also holds for $-\frac{\pi}{2} < x < 0$, as all three functions in the inequality are even.
Since $\cos$ is continuous at $0$, and $t\mapsto t^2$ is continuous at $\cos(0)=1$, the function $x\mapsto \cos^2(x)$ is continuous at $0$, i.e. $$ \lim_{x \to 0}\cos^2(x)=1 \, . $$ Therefore, by the squeeze theorem, $$\lim_{x \to 0}\frac{\sin(x)}{x}=1 \, .$$
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Here is a slick trick using elementary integration methods. Note that \begin{align} \int_0^1 \ \cos(xt) \ dt & = \left[ \dfrac{1}{x} \cdot\sin(xt) \right]_0^1 \\ & =\dfrac{\sin (x)}{x} - \dfrac{\sin (0)}{x} \\ & = \dfrac{\sin (x)}{x}. \end{align} Hence, \begin{align} \lim_{x \rightarrow 0} \dfrac{\sin (x)}{x} & = \lim_{x \rightarrow 0} \int _0^1 \ \cos(xt) \ dt \\ & = \int_0^1 \cos (0) \ dt \\ & =1. \end{align}
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This geometric solution comes form this question according to the following sketch
we have
$$Area(OBP) \le Area(OAP)\le Area(OBP)+Area(ABPQ)$$
that is
$$\frac 12 \cos x |\sin x|\le \frac12 \cdot 1 \cdot |x|\le \frac 12 \cos x |\sin x|+(1-\cos x)|\sin x|=|\sin x|- \frac12 \cos x |\sin x|$$
and dividing by $\frac12|\sin x|>0$
$$ \cos x \le \frac{|x|}{|\sin x|}\le 2- \cos x $$
and since $\frac{|x|}{|\sin x|} =\frac{x}{\sin x}>0$ for $x\neq 0$ we obtain
$$ \cos x \le \frac{x}{\sin x}\le 2- \cos x $$
finally by squeeze theorem since
we conclude that
$$\lim_{x\to 0}\frac x{\sin x}=1$$
The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$ \frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1} $$ Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get $$ \cos(x)\le\frac{\sin(x)}{x}\le1\tag{2} $$ Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3} $$ Also, dividing $(2)$ by $\cos(x)$, we get that $$ 1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4} $$ Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5} $$