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

###### Theorem 1.
 $\lim_{x\to 0}\frac{\sin x}{x}=1$

$x<\tan x$ for $\displaystyle 0.

###### Proof.

First, let $\displaystyle 0. Then $0<\cos x<1$. Note also that

 $x<\tan x.$ (1)
 $x\cos x<\sin x.$ (2)

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

 $\sin x (3)

Combining inequalities (2) and (3) gives

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

Dividing by $x$ yields

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

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

 $\cos(-x)<\frac{\sin(-x)}{-x}<1.$ (6)
 $\cos x<\frac{-\sin x}{-x}<1.$ (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 LimitOfdisplaystylefracsinXxAsXApproaches0 2013-03-22 16:58:45 2013-03-22 16:58:45 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Theorem msc 26A06 msc 26A03 ComparisonOfSinThetaAndThetaNearTheta0 SincFunction DerivativesOfSinXAndCosX DerivativesOfSineAndCosine