limit of sinxx as x approaches 0

Theorem 1.

x<tanx for 0<x<π2.


First, let 0<x<π2. Then 0<cosx<1. Note also that

x<tanx. (1)

Multiplying both of this inequalityMathworldPlanetmath by cosx yields

xcosx<sinx. (2)

By this theorem (,

sinx<x. (3)

Combining inequalities (2) and (3) gives

xcosx<sinx<x. (4)

Dividing by x yields

cosx<sinxx<1. (5)

Now let -π2<x<0. Then 0<-x<π2. Plugging -x into inequality (5) gives

cos(-x)<sin(-x)-x<1. (6)

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

cosx<-sinx-x<1. (7)

Therefore, inequality (5) holds for all real x with 0<|x|<π2.

Since cos is continuousMathworldPlanetmath, limx0cosx=cos0=1. Thus,

1=limx0cosxlimx0sinxxlimx01=1. (8)

By the squeeze theorem, it follows that limx0sinxx=1. ∎

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

Title limit of sinxx 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