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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.

  • 2
    Sandwich theorem may be applied to prove it.2017-07-04

25 Answers 25

525

sinc and tanc at 0

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} $

  • 0
    @AbdelrhmanFawzy: so that we know that $\lim\limits_{x\to0}\cos(x)=\cos(0)=1$.2018-08-25
128

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.

diagram

  • 0
    Sir which software/website have you used to make this diagram?2017-03-17
98

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.

  • 2
    @Sven : That is indeed how it's done in Robinson's "nonstandard analysis". There is another approach to rigorous infinitesimals in which it would be done the way Euler did it, saying that if $\theta$ is infinitely small then $\dfrac{\sin\theta}\theta = 1.$ It's called "smooth infinitesimal analysis". In that approach, the square of an infinitesimal is $0$, so we have $ \frac{\sin\theta} \theta = \frac{\theta - \dfrac{\theta^3} 6 + \dfrac{\theta^5}{120} - \cdots \cdots} \theta = 1. $ https://en.wikipedia.org/wiki/Smooth_infinitesimal_analysis $\qquad$2017-04-14
79

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:

Copy of the picture from the Fatos Matematicos blog

  • 1
    @Dr. MV: if you rescale the circle in $\cos \theta$ units, then the original $\sin \theta$ is a scaled $\tan \theta$ which is always longer than its arc.2016-05-30
52

I claim that for $0 that the following holds $\sin x \lt x \lt \tan x$ Figure 1
In the diagram, we let $OC=OA=1$. In other words, $Arc\:CA=x$ is an arc of a unit circle. The shortest distance from point $C$ to line $AO$ is line $CE=\sin x$ (because $CE\perp OA$). Another path from point $C$ to line $OA$ is arc $CA$ (which is longer than CE because it is not the shortest path). So we have at the very least $\sin x \lt x$ Now we need to show that line $BA=\tan x \gt x$.
Lines $AD$ and $CD$ are both tangent to arc $CA$. $CD+DA$ is longer than arc $CA$ because the set of points bound by sector $OCA$ is a subset of the set of points bound by quadrilateral $OCDA$, both of which are convex sets. This means that the perimeter of quadrilateral $OCDA$ must be longer than the perimeter of the sector $OCA$ (as per Archimedes, On the Sphere and Cylinder Book I). But both the sector and the quadrilateral both have sides $OC$ and $OA$, so we have $CA=x But $BD>CD$ because it is the hypotenuse in $\triangle BCD$ we have $\tan x = BA = BD+DA\gt CD+DA \gt CA=x \gt \sin x$

So we have $\sin x \lt x \lt \tan x$ $\frac{\sin x}{x} \lt 1 \lt \frac{\tan x}{x}=\frac{\sin x}{x}\cdot\sec x$ From this we can extract $\frac{\sin x}{x} \lt 1$ and $1 \lt \frac{\sin x}{x}\cdot\sec x$ $\cos x \lt \frac{\sin x}{x}$ Putting these inequalities back together we have $\cos x \lt \frac{\sin x}{x} \lt 1$

Because $\displaystyle\lim_{x\to 0}\cos x = 1$, by the squeeze theorem we have $\lim_{x\to 0}\frac{\sin x}{x}=1$

  • 2
    If you don't have a copy of Archimedes's book, you can approximate sector OCA with a polygon inside OCDA and use http://www.cut-the-knot.org/m/Geometry/PerimetersOfTwoConvexPolygons.shtml2017-06-06
51

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.

  • 2
    this assumes the *a priori* knowledge of the existence of the limit $\lim_{x\to 0}\frac{\sin(x)}{x}$, how would you go about proving it before hand? (+1) by the way for the original and alternative proof2016-02-24
48

I am not sure if it counts as proof, but I have seen this done by a High Schooler. enter image description here

In the given picture above, $\displaystyle 2n \text{ EJ} = 2nR \sin\left( \frac{\pi}{n } \right ) = \text{ perimeter of polygon }$.

$\displaystyle \lim_{n\to \infty }2nR \sin\left( \frac{\pi}{n } \right ) = \lim_{n\to \infty } (\text{ perimeter of polygon }) = 2 \pi R \implies \lim_{n\to \infty}\frac{\sin\left( \frac{\pi}{n } \right )}{\left( \frac{\pi}{n } \right )} = 1$ and let $\frac{\pi}{n} = x$.

  • 0
    @Imago http://math.stackexchange.com/questions/720935/historic-proof-of-the-area-of-a-circle/1678093#16780932016-06-06
43

Don't you feel strange about why most of the proofs are done with a figure? I've had this problem in the beginning, and realized after that this is due to the definition we use for the function $\sin x$. Because the usual definition of $\sin x$ we all study first in high schools depends on “classical geometry” and usually with a figure, you should depict out the figure and to make it clear.

However, if you use other definitions of $\sin x$ that are equivalent to the former, you'll find it more simple. For example,

$\sin x = \frac{x^1}{1!} - \frac{x^3}{3!}+ \frac{x^5}{5!} - \cdots + \cdots - \cdots$

and hence

$\frac{\sin x}x = \frac{x^0}{1!} - \frac{x^2}{3!}+ \frac{x^4}{5!} - \cdots$

which obviously tends to $1$ as $x$ approaches 0.

  • 0
    @celschk starting with the infinite series as a 'definition' causes more problems than it solves. Now we have to prove that this sine behaves like the sine we learned in high school. This proof is compelling but it is not really a proof.2016-09-24
30

Here's one more: $ \lim_{x \to 0} \frac{\sin x}{x}=\lim_{x \to 0} \lim_{v \to 0}\frac{\sin (x+v)-\sin v}{x}\\ =\lim_{v \to 0} \lim_{x \to 0}\frac{\sin (x+v)-\sin v}{x}=\lim_{v \to 0}\sin'v=\lim_{v\ \to 0} \cos v=1 $

  • 32
    Usually, this limit is used to compute the derivative of $\sin(x)$.2013-07-20
28

It depends on your definition of the sine function. I would suggest checking out the geometric proof in ProofWiki.

23

The strategy is to find $\frac{d\arcsin y}{dy}$ first. This can easily be done using the picture below.

arcsin as area

From the above picture, $\arcsin y$ is twice the area of the orange bit. The area of the red bit is ${1 \over 2}y\sqrt{1-y^2}$. The area of the red bit plus the orange bit is $\int_{0}^y \sqrt{1-Y^2} dY$. So $\arcsin y = 2\int_{0}^y \sqrt{1-Y^2} dY - y\sqrt{1-y^2}.$ Differentiating with respect to $y$ gives $\frac{d\arcsin y}{dy} = \frac{1}{\sqrt{1-y^2}}$. Using the theorem for the derivative of inverse functions yields $\sin' \theta = \sqrt{1 - \sin^2 \theta} = \cos \theta$.

(A similar thing can be done with the arc length definition of $\arcsin$.)

  • 0
    While this is indeed an interesting approach, integrals haven't been taught at the point where this limit is proved. Thank you for your answer though.2015-04-11
21

Let $f:\{y\in\mathbb{R}:y\neq 0\}\to\mathbb{R}$ be a function defined by $f(x):=\dfrac{\sin x}{x}$ for all $x\in \{y\in\mathbb{R}:y\neq 0\}$.

We have $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x}=1$ if and only if for every $\varepsilon>0$, there exists a $\delta>0$ such that $|f(x)-1|<\varepsilon$ whenever $0<|x-0|<\delta$.

Let $\varepsilon>0$ be an arbitrary real number.

Note that $\sin x=\displaystyle \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$.

If $x \neq 0$, we have $\dfrac{\sin x}{x}=$\displaystyle \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n+1)!}=1+$\displaystyle \sum_{n=1}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n+1)!}$.

We thus have

$|f(x)-1|=\left|\dfrac{\sin x}{x}-1\right|=\left|\displaystyle \sum_{n=1}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n+1)!}\right|\leq \left|\displaystyle\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n+1)!}\right|\leq \displaystyle\sum_{n=1}^{\infty} \left|\dfrac{x^{2n}}{(2n+1)!}\right|$

Therefore we have

$|f(x)-1|\leq \displaystyle\sum_{n=1}^{\infty} \left|\dfrac{x^{2n}}{(2n+1)!}\right|\leq \displaystyle \sum_{n=1}^{\infty} |x^{2n}|=\sum_{n=1}^{\infty}|x^2|^n$

If $0<|x|<1$, then $0<|x^2|<1$, and the infinite series $\displaystyle\sum_{n=1}^{\infty}|x^2|^n$ converges to $\dfrac{x^2}{1-x^2}$.

Choose $\delta:=\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}$. Then $0<|x-0|<\delta$ implies that $0<|x|<$$\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}<1$, and hence $x^2<\varepsilon-\varepsilon x^2$. But $x^2<\varepsilon-\varepsilon x^2$ implies that $\dfrac{x^2}{1-x^2}<\varepsilon$.

We therefore have $\sum_{n=1}^{\infty}|x^2|^n<\varepsilon$ whenever $0<|x-0|<\delta$. But since $|f(x)-1|\leq\displaystyle\sum_{n=1}^{\infty}|x^2|^n$, we have $|f(x)-1|<\varepsilon$ whenever $0<|x-0|<\delta$.

Since $\varepsilon$ was arbitrary, we have $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x}=1$.

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    The premise of the question was not to use power series.2018-12-04
19

Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest without appeal to geometry or differential calculus.

Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.


We define the sine function, $\sin(x)$, as the inverse function of the function $f(x)$ given by

$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt }\tag 1$

for $|x|< 1$.

NOTE: It can be shown that the sine function defined as the inverse of $f(x)$ given in $(1)$ has all of the familiar properties that characterize the circular function $\sin(x)$.

It is straightforward to show that since $\frac{1}{\sqrt{1-t^2}}$ is positive and continuous for $t\in (-1,1)$, $f(x)$ is continuous and strictly increasing for $x\in (-1,1)$ with $\displaystyle\lim_{x\to 0}f(x)=f(0)=0$.

Therefore, since $f$ is continuous and strictly increasing, its inverse function, $\sin(x)$, exists and is also continuous and strictly increasing with $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.


From $(1)$, we have the bounds (SEE HERE)

$\bbox[5px,border:2px solid #C0A000]{1 \le \frac{f(x)}x\le \frac{1}{\sqrt{1-x^2}}} \tag 2$

for $x\in (-1,1)$, whence applying the squeeze theorem to $(2)$ yields

$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$


Finally, let $y=f(x)$ so that $x=\sin(y)$. As $x\to 0$, $y\to 0$ and we can write $(3)$ as

$\lim_{y\to 0}\frac{y}{\sin(y)}=1$

from which we have

$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0}\frac{\sin(y)}{y}=1}$

as was to be shown!


NOTE:

We can deduce the following set of useful inequalities from $(2)$. We let $x=\sin(\theta)$ and restrict $x$ so that $x\in [0,1)$. In addition, we define new functions, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$ and $\tan(\theta)=\sin(\theta)/\cos(\theta)$.

Then, we have from $(2)$

$\bbox[5px,border:2px solid #C0A000]{y\cos(y)\le \sin(y)\le y\le \tan(y)} $

which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.

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    @ogogmad "Abstract Definitions?" Define that term. These are "Rigorous Definitions." And the squeeze theorem is completely independent of the FOC. It is, therefore, more general in this application. You are free to modify your answer as you see fit, of course.2018-02-05
15

Because $\sin x$ has zeroes at $x=n \pi$ for arbitrary integer $n$ including $x=0$, you can use Vieta's Theorem $\sin x = A(\cdots(x+2 \pi)(x+\pi)x(x-\pi)(x-2 \pi)\cdots)$ with a constant $A$. Because $\sin(\frac{\pi}{2})=1$ this constant can be determined by the equation: $1=A(\cdots(\frac{\pi}{2}+2 \pi)(\frac{\pi}{2}+\pi)\frac{\pi}{2}(\frac{\pi}{2}-\pi)(\frac{\pi}{2}-2 \pi)\cdots).$

Now, in the Expression $f(x):= \frac{\sin(x)}{x}$ the $x$ cancels such that $f(x)=A(\cdots(x+2 \pi)(x+\pi)(x-\pi)(x-2 \pi)\cdots),$ hence: $\lim_{x \rightarrow 0} f(x)=A(\cdots(2 \pi) \cdot \pi\cdot(- \pi)\cdot(-2 \pi)\cdots) = A \prod_{k=1}^\infty (-k^2 \pi^2).$

$\frac{1}{A} = \frac{\pi}{2} \prod_{k=1}^\infty ((\frac{\pi}{2})^2-k^2 \pi^2)$. The proof is completed when the Wallis product is used.

  • 2
    Ontop of Antonio Vargas comment, Vieta's theorem only works for finite number of zeros. You need the Weierstrass-Factorization Theorem or Hadamard-Factorization Theorem. Btw you also need to verify that $\sin(x)$ does not have any complex zeros.2016-06-10
15

Usual proofs can be circular, but there is a simple way for proving such inequality.

Let $\theta$ be an acute angle and let $O,A,B,C,D,C'$ as in the following diagram: enter image description here

We may show that:

$ CD \stackrel{(1)}{ \geq }\;\stackrel{\large\frown}{CB}\; \stackrel{(2)}{\geq } CB\,\stackrel{(3)}{\geq} AB $

$(1)$: The quadrilateral $OCDC'$ and the circle sector delimited by $O,C,C'$ are two convex sets. Since the circle sector is a subset of the quadrilateral, the perimeter of the circle sector is less than the perimeter of the quadrilateral.

$(2)$: the $CB$ segment is the shortest path between $B$ and $C$.

$(3)$ $CAB$ is a right triangle, hence $CB\geq AB$ by the Pythagorean theorem.

In terms of $\theta$ we get: $ \tan\theta \geq \theta \geq 2\sin\frac{\theta}{2} \geq \sin\theta $ for any $\theta\in\left[0,\frac{\pi}{2}\right)$. Since the involved functions are odd functions the reverse inequality holds over $\left(-\frac{\pi}{2},0\right]$, and $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$ follows by squeezing.


A slightly different approach might be the following one: let us assume $\theta\in\left(0,\frac{\pi}{2}\right)$.
By $(2)$ and $(3)$ we have $ \theta \geq 2\sin\frac{\theta}{2}\geq \sin\theta $ hence the sequence $\{a_n\}_{n\geq 0}$ defined by $a_n = 2^n \sin\frac{\theta}{2^n}$ is increasing and bounded by $\theta$. Any increasing and bounded sequence is convergent, and we actually have $\lim_{n\to +\infty}a_n=\theta$ since $\stackrel{\large\frown}{BC}$ is a rectifiable curve and for every $n\geq 1$ the $a_n$ term is the length of a polygonal approximation of $\stackrel{\large\frown}{BC}$ through $2^{n-1}$ equal segments. In particular

$ \forall \theta\in\left(0,\frac{\pi}{2}\right), \qquad \lim_{n\to +\infty}\frac{\sin\left(\frac{\theta}{2^n}\right)}{\frac{\theta}{2^n}} = 1 $ and this grants that if the limit $\lim_{x\to 0}\frac{\sin x}{x}$ exists, it is $1$. By $\sin x\leq x$ we get $\limsup_{x\to 0}\frac{\sin x}{x}\leq 1$, hence it is enough to show that $\liminf_{x\to 0}\frac{\sin x}{x}\geq 1$. We already know that for any $x$ close enough to the origin the sequence $\frac{\sin x}{x},\frac{\sin(x/2)}{x/2},\frac{\sin(x/4)}{x/4},\ldots$ is convergent to $1$, hence we are done.

Long story short: $\lim_{x\to 0}\frac{\sin x}{x}=1$ follows from the fact that a circle is a rectifiable curve, and a circle is a rectifiable curve because it is the boundary of a convex, bounded subset of $\mathbb{R}^2$. The convexity of the disk follows from the triangle inequality: a disk is a closed ball for the euclidean distance.


$(1)$ relies on this powerful Lemma:

Lemma. If $A,B$ are convex bounded sets in $\mathbb{R}^2$ and $A\subsetneq B$, the perimeter of $A$ is less than the perimeter of $B$.

Proof: by boundedness and convexity, $\partial A$ and $\partial B$ are rectifiable, with lengths $L(A)=\mu(\partial A),\,L(B)=\mu(\partial B)$. Always by convexity, there is some chord in $B$ that does not meet the interior of $A$ (a tangent to $\partial A$ at a smooth point does the job, for instance). Assume that such chord has endpoints $B_1, B_2 \in \partial B$ and perform a cut along $B_1 B_2$: both the area and the perimeter of $B$ decrease, but $B$ remains a bounded convex set enclosing $A$. Since $A$ can be approximated through a sequence of consecutive cuts, $L(A) follows.

  • 0
    Great answer (+1)2018-01-25
12

Let $\sin(x)$ is defined as solution of $\frac{d^2}{dx^2}\textrm{f}(x)=-\textrm{f}(x)$ with $\mathrm f(0)=0,\,\frac{d}{dx}\mathrm f(0)=C$ initial conditions, so exact solution is $\mathrm f(x)=C\cdot\sin(x)$. Define second derivative as $ \begin{align*} \frac{d^2}{dx^2}\textrm{f}(x)=\lim_{\Delta x\to 0}{\frac{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}-\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}{\Delta x}}&=\\=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-2\cdot \mathrm f(x-\Delta x)+\mathrm f(x-2\cdot\Delta x)}{\Delta x^2}} \end{align*} $ we can easy check this limit for any (?) functions. Similarly, we can define the first derivative for right, middle and left points: $ \frac{d}{dx}\textrm{f}(x)\left\{ \begin{aligned} &=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}} \\ &=\lim_{\Delta x\to 0}{\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}\\ &=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-2\cdot\Delta x)}{2\cdot\Delta x}} \end{aligned} \right. $ Let's use the finite elements method assuming $Td=\Delta x,\,y_n=\mathrm f(x),\,y_{n-1}=\mathrm f(x-\Delta x),\,y_{n-2}=\mathrm f(x-2\cdot \Delta x)$ Override differential equation as $ \frac{y_n-2\cdot y_{n-1}+y_{n-2}}{Td^2}=-y_n $ Now solve this implicit equation for $y_n$ to obtain explicit recurrence relation: $ y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2} $ Using arbitrarily small but non-zero quantity Td we can plot exponentially decaying sampled sine function (because the poles are inside the unit circle of the transfer function corresponding to the given recurrence relation). Similarly we write three systems for the initial conditions:

$ \left\{ \begin{aligned} &y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2} \\ &C=\frac{y_n-y_{n-1}}{Td} \end{aligned}\right. $ $ \left\{ \begin{aligned} &y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2} \\ &C=\frac{y_{n-1}-y_{n-2}}{Td} \end{aligned}\right. $ $ \left\{ \begin{aligned} &y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2} \\ &C=\frac{y_n-y_{n-2}}{2\cdot Td} \end{aligned}\right. $ Solve this sequence of equations for $y_{n-1}$ and $y_{n-2}$: $ \left\{ \begin{aligned} &y_{n-1} = -C\cdot Td + y_{n}\\ &y_{n-2}=-2\cdot C\cdot Td + y_{n}\cdot\left(1-Td^2\right)\ \end{aligned}\right. $ $ \left\{ \begin{aligned} &y_{n-1} = -C\cdot Td + y_{n}\cdot\left(1+Td^2\right)\\ &y_{n-2}=-2\cdot C\cdot Td + y_{n}\cdot\left(1+Td^2\right)\ \end{aligned}\right. $ $ \left\{ \begin{aligned} &y_{n-1} = -C\cdot Td + y_{n}\cdot\left(1+\frac{Td^2}{2}\right)\\ &y_{n-2}=-2\cdot C\cdot Td + y_{n}\ \end{aligned}\right. $ At zero point $y_n=\mathrm f(0)=0$ and we can see linear dependence: $ \begin{aligned} &y_{n-1} = -C\cdot Td\\ &y_{n-2}=-2\cdot C\cdot Td \end{aligned} $ for all three solutions. Replace back: $ \begin{array}{l} \mathrm f(0)&=0\\ \mathrm f(0-\Delta x) &= -C\cdot \Delta x\\ \mathrm f(0-2\cdot \Delta x) &= -2\cdot C\cdot \Delta x \end{array} $ So all three $\frac{d}{dx}\mathrm f(0)$ limits is equal to $C$ at $x=0$ and in accordance with $\mathrm f(x)=C\cdot\sin(x)$ by definition we can write $ \lim_{\Delta x\to 0}{\frac{\mathrm f(0)-\mathrm f(0-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{0-(-C \cdot \Delta x)}{\Delta x}}=C $ Thus $ \lim_{\Delta x\to 0}{\frac{\sin(0)-C\cdot\sin(0-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{C\cdot\sin(\Delta x)}{\Delta x}}=C\cdot\lim_{\Delta x\to 0}{\frac{\sin(\Delta x)}{\Delta x}}=C $ and $\lim_{\Delta x\to 0}{\frac{\sin(\Delta x)}{\Delta x}}=1$

  • 0
    @FUZxxl do you realize that users with different levels and mathematical background may check this question? The answer given above might be helpful for them.2017-09-23
10

Simple one is using sandwich theorem Which demonstrated earlier.In this method you have to show that $\frac{\sin x}{x} $ lies between other two functions. As $x \longrightarrow 0$ both of them will tends to ONE.

Then as in the case of sandwich (if both the bread part go to one stomach the middle portion will also go to the same stomach) $\frac{\sin x}{x}$ will go to ONE.

You can use geogebra to see the visualization of this phenomena using geogebra.First you input $\sin x$ and $x$ and observe that near to $0$ values of $\sin x$ and $x$ are same.

Secondly input $\frac{\sin x}{x}$ then observe function is approaching to $1$ as $x$ tends to $0$

  • 4
    This describes the Sandwich Theorem, but does not answer the question. At best, this should be a comment to the question,2014-12-13
8

Originally posted on the proofs without words post, here is a simple image that explains the derivative of $\sin(x)$, which as we all know, is directly related to the limit at hand.

enter image description here

If one is not so convinced, take a look at the above picture and notice that if $u\pm h$ is in the first quadrant, then

$\frac{\sin(x+h)-\sin(x)}h<\cos(x)<\frac{\sin(x-h)-\sin(x)}{-h}$


Notice that

$ \begin{align}\frac{d}{dx}\sin(x)&=\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}h\\\text{picture above}&=\lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}h\\\cos(x)&=\lim_{h\to0}\sin(x)\frac{\cos(h)-1}h+\cos(x)\frac{\sin(h)}h\\\cos(0)&=\lim_{h\to0}\frac{\sin(h)}h\end{align} $

  • 0
    @FUZxxl Since you still seem unhappy, I've added a better "squeeze theorem" type thing for the derivative of $\sin(x)$.2017-01-05
5

Here is another approach.

enter image description here(1) picture 2(2)

In the large triangle, $\tan(\theta)=\frac{opp}{adg}=\frac{z}{1}=z$ So the triangle has height $z=\tan(\theta)$ and base $1$ so it's area is $Area(big triangle)=\frac{1}{2}z=\frac{1}{2}\tan(\theta)$

Next the sector area as a fraction of the entire circle, the sector is (see the right hand side of picture (1))$\frac{\theta}{2\pi}$ of the entire circle so it's area is

$Area(sector)=\frac{\theta}{2\pi}*(\pi)(1)^2=\frac{\theta}{2}$ The triangle within the sector has height $y$. But $y=\frac{y}{1}=\frac{opp}{hyp}=\sin(\theta)$ so the small triangle has height $y=\sin(\theta)$ and base $1$ so its area is $Area(small triangle)=\frac{1}{2}y=\frac{1}{2}\sin(\theta)$ Now we can use the sandwich theorem as $Area(big triangle)\geq Area(sector)\geq Area(small triangle)$

using the equations we worked out above this becomes

$\frac{\tan(\theta)}{2}\geq\frac{\theta}{2}\geq\frac{\sin(\theta)}{2}$ now multipliying through by two and using the fact that $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$ we get that $\frac{\sin(\theta)}{\cos(\theta)}\geq\theta\geq\sin(\theta)$ taking reciprocals changes the inequalities so we have that $\frac{\cos(\theta)}{\sin(\theta)}\leq\frac{1}{\theta}\leq\frac{1}{\sin(\theta)}$ now finally multiplying through by $\sin(\theta)$ we get $\cos(\theta)\leq\frac{\sin(\theta)}{\theta}\leq1$ Now $\lim \limits_{\theta \to 0}\cos(\theta)=1$ and$\lim \limits_{\theta \to 0}1=1$

so by the sandwhich theorem $\lim \limits_{\theta \to 0}\frac{\sin(\theta)}{\theta}=1$ also. QED

4

We may use a linear approximation for this limit. The following expression comes straight out of the equation of a line that is $y=mx+b$. Using the fact that $m=\dfrac{d}{dx}[f(x)]$ $f(x) \approx f(a) + f'(a)(x-a)$In this case $f(x)=\sin x$ and $a=0$ $\implies \sin x \approx f(0)+\cos 0 (x-0)=x \longleftrightarrow \sin x \approx x$. The following graph better illustrates this tangent line approximation.

Since, $\sin x \approx x \implies \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}=1$. Quod Erat Demonstrandum

2

Here is a proof to those familiar with power series.

The definition of $\sin(x)$ is

$\sin(x) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}h^{2k+1}$

Therefore we get

$\begin{align} \lim_{x \to 0} \frac{\sin(x)}{x} &= \lim_{x \to 0} \frac{\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1}}{x} \\&= \lim_{x \to 0} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k} \\&= 1 + \lim_{x \to 0} \sum_{k=1}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k} \\&= 1 \end{align}$

where we have used the fact that the power series $\sum_{k=1}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k}$ has radius of convergence $R=\infty$ and therefore is continuous on $\mathbb R$. This allows us to take the limit inside and we get

$\lim_{x \to 0} \sum_{k=1}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k} = \sum_{k=1}^\infty \frac{(-1)^k}{(2k+1)!} 0^{2k} = 0$

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    The premise of the question is not to use power series.2018-12-04