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Answer by user for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
According to the following sketch:using the properties for the isosceles triangle, the area of a circular sector is given by the limit:$$\frac{\theta}2 \cdot 1^2 = \lim_{N\to \infty} N\cdot...
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Answer by user for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
This geometric solution comes form this question according to the following sketchwe have$$Area(OBP) \le Area(OAP)\le Area(OBP)+Area(ABPQ)$$that is$$\frac 12 \cos x |\sin x|\le \frac12 \cdot 1 \cdot...
View ArticleAnswer by Bananas for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
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} -...
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Answer by Joe for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
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...
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Answer by SurfaceIntegral for How to prove that...
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...
View ArticleAnswer by zkutch for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
For completeness answers let me suggest axiomatic approach to $\sin$ and $\cos$. One possible definition is here. I find another one(Ilyin, Poznyak: Fundamentals of Mathematical Analysis, 2005, vol.1,...
View ArticleAnswer by ChoMedit for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
How about this proof?We can check that function defined as\begin{align}\int_{-m}^m e^{2\pi i k x} \mathrm{d}{k}\end{align} is continuous and have a value $2m$ at $x=0$.It is same...
View ArticleAnswer by Behnam Esmayli for How to prove that $\lim\limits_{x\to0}\frac{\sin...
The answer ultimately depends on how you define $\sin x$ in the first place.Here is a more fun one! $\sin x$ is the unique function satisfying $$ y'' = -y; y(0)=0, y'(0)=1 $$By Theory of Ordinary...
View ArticleAnswer by The_Sympathizer for How to prove that...
This is a new post on an old saw because this is one of those things where that I can see how that, all too sadly, the way in which we've structured the current maths curriculum really doesn't make it...
View ArticleAnswer by Archer for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
We can also use Euler's formula to prove the limit: $$e^{ix} = \cos x + i\sin x$$$$\lim_{x\to 0}\dfrac{\sin x}{x} = \implies \lim_{x\to 0} \dfrac{e^{ix}- e^{-ix}}{2i x}$$$$= \lim_{x\to 0}...
View ArticleAnswer by mdcq for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
Here is a proof to those familiar with power series.The definition of $\sin(x)$ is$$\sin(x) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}h^{2k+1}$$Therefore we get$$\begin{align} \lim_{x \to 0}...
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Answer by Jack D'Aurizio for How to prove that $\lim\limits_{x\to0}\frac{\sin...
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:We may show that:$$ CD...
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Answer by user395952 for How to prove that $\lim\limits_{x\to0}\frac{\sin...
Here is another approach.(1)(2)In the large triangle, $$\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...
View ArticleAnswer by Mark Viola for How to prove that $\lim\limits_{x\to0}\frac{\sin...
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...
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Answer by Simply Beautiful Art for How to prove that...
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.If one is not so...
View ArticleAnswer by Supreeth Narasimhaswamy for How to prove that...
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...
View ArticleAnswer by Timur Zhoraev for How to prove that $\lim\limits_{x\to0}\frac{\sin...
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...
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Answer by wlad for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
The strategy is to find $\frac{d\arcsin y}{dy}$ first. This can easily be done using the picture below.From the above picture, $\arcsin y$ is twice the area of the orange bit. The area of the red bit...
View ArticleAnswer by user223261 for How to prove that $\lim\limits_{x\to0}\frac{\sin...
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...
View ArticleAnswer by Madhu for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
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...
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Answer by John Joy for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
I claim that for $0<x<\pi/2$ that the following holds$$\sin x \lt x \lt \tan x$$In the diagram, we let $OC=OA=1$. In other words, $Arc\:CA=x$ is an arc of a unit circle. The shortest distance...
View ArticleAnswer by Alex for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
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...
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Answer by S L for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
I am not sure if it counts as proof, but I have seen this done by a High Schooler.In the given picture above, $\displaystyle 2n \text{ EJ} = 2nR \sin\left( \frac{\pi}{n } \right ) = \text{ perimeter of...
View ArticleAnswer by user 1591719 for How to prove that $\lim\limits_{x\to0}\frac{\sin...
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}...
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Answer by Paulo Sérgio for How to prove that $\lim\limits_{x\to0}\frac{\sin...
Look at this link:http://fatosmatematicos.blogspot.com/2010/08/provas-sem-palavras-parte-20.htmlHere is the picture I copied from that blog:
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Answer by robjohn for How to prove that $\lim\limits_{x\to0}\frac{\sin x}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...
View ArticleAnswer by Michael Hardy for How to prove that $\lim\limits_{x\to0}\frac{\sin...
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...
View ArticleAnswer by Yuval Filmus for How to prove that $\lim\limits_{x\to0}\frac{\sin...
It depends on your definition of the sine function. I would suggest checking out the geometric proof in ProofWiki.
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Answer by tkr for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
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$$and rearranging $1 < {\tan x \over x} = {\sin x \over x...
View ArticleHow to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
How can one prove the statement$$\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...
View ArticleAnswer by peter a g for How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
Not at all appropriate for a cal one course (or any course), but for the perverse, egregious silliness of it:A circle is parametrized by $\gamma(t) = (\cos t, \sin t)$. Now, geometrically, the tangent...
View ArticleAnswer by Cyclonestopper for How to prove that $\lim\limits_{x\to0}\frac{\sin...
A well-known calculus rule is L'hopital's rule, which states that if the numerator and denominator of a limit are both 0, then you can take the derivative of the numerator and denominator and try to...
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