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limit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0 (Theorem)
Theorem  
$\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$

Note that this entry uses the result $ x<\tan x$ for $ \displaystyle 0<x<\frac{\pi}{2}$. I have received word that a proof of this fact which uses no calculus will be supplied on PlanetMath. Please let me know when this is completed so that I can edit this entry accordingly.

Proof. First, let $ \displaystyle 0<x<\frac{\pi}{2}$. Then $ 0<\cos x<1$. Note also that
$\displaystyle x<\tan x.$ (1)

Multiplying both sides of this inequality by $ \cos x$ yields

$\displaystyle x\cos x<\sin x.$ (2)

By this theorem,

$\displaystyle \sin x<x.$ (3)

Combining inequalities (2) and (3) gives

$\displaystyle x\cos x<\sin x<x.$ (4)

Dividing by $ x$ yields

$\displaystyle \cos x<\frac{\sin x}{x}<1.$ (5)

Now let $ \displaystyle \frac{-\pi}{2}<x<0$. Then $ \displaystyle 0<-x<\frac{\pi}{2}$. Plugging $ -x$ into inequality (5) gives

$\displaystyle \cos (-x)<\frac{\sin (-x)}{-x}<1.$ (6)

Since $ \cos$ is an even function and $ \sin$ is an odd function, we have

$\displaystyle \cos x<\frac{-\sin x}{-x}<1.$ (7)

Therefore, inequality (5) holds for all real $ x$ with $ \displaystyle 0<\vert x\vert<\frac{\pi}{2}$.

Since $ \cos$ is continuous, $ \displaystyle \lim_{x \to 0} \cos x=\cos 0=1$. Thus,

$\displaystyle 1=\lim_{x \to 0} \cos x \le \lim_{x \to 0} \frac{\sin x}{x} \le \lim_{x \to 0} 1=1.$ (8)

By the squeeze theorem, it follows that $ \displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$. $ \qedsymbol$

Note that the above limit is also valid if $ x$ is considered as a complex variable.



"limit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0" is owned by Wkbj79.
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See Also: comparison of $\sin þeta$ and $þeta$ near $þeta = 0$, sinc function, derivatives of $\sin x$ and $\cos x$, derivatives of sine and cosine


Attachments:
limit of $\displaystyle \frac{1-\cos x}{x}$ as $x$ approaches 0 (Corollary) by Wkbj79
limit examples (Example) by pahio
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Cross-references: variable, complex, limit, squeeze theorem, continuous, real, odd function, even function, inequality
There are 7 references to this entry.

This is version 7 of limit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0, born on 2007-04-24, modified 2007-08-18.
Object id is 9255, canonical name is LimitOfDisplaystyleFracsinXxAsXApproaches0.
Accessed 1968 times total.

Classification:
AMS MSC: 26A03 (Real functions :: Functions of one variable :: Foundations: limits and generalizations, elementary topology of the line)
 26A06 (Real functions :: Functions of one variable :: One-variable calculus)
Pending Errata and Addenda
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Simpler proof by kfgauss70 on 2008-01-02 12:28:14
Since sin(x)/x is an even function, the proof for the theorem could be made more elegant just considering the case x>0. In other words, it would suffice to prove that the limit is 0 for x->0+.
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