limit of $\displaystyle\frac{\mathop{sin}\nolimits x}{x}$ as $x$ approaches 0

\lim_{x\to 0}\frac{\sin x}{x}=1

$x<\tan x$ for $\displaystyle 0<x<\frac{\pi}{2}$ .

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

x<\tan x.

(1)

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

x\cos x<\sin x.

(2)

By this theorem (http://planetmath.org/ComparisonOfSinThetaAndThetaNearTheta0),

\sin x<x.

(3)

Combining inequalities (2) and (3) gives

x\cos x<\sin x<x.

(4)

Dividing by $x$ yields

\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

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

(6)

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

\cos x<\frac{-\sin x}{-x}<1.

(7)

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

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

1=\lim_{x\to 0}\cos x\leq\lim_{x\to 0}\frac{\sin x}{x}\leq\lim_{x\to 0}1=1.

(8)

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

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

Title	limit of $\displaystyle\frac{\mathop{sin}\nolimits x}{x}$ as $x$ approaches 0
Canonical name	LimitOfdisplaystylefracsinXxAsXApproaches0
Date of creation	2013-03-22 16:58:45
Last modified on	2013-03-22 16:58:45
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	10
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 26A06
Classification	msc 26A03
Related topic	ComparisonOfSinThetaAndThetaNearTheta0
Related topic	SincFunction
Related topic	DerivativesOfSinXAndCosX
Related topic	DerivativesOfSineAndCosine

limit of s⁢i⁢nxx as x approaches 0