How do you find the limit of #(sinx-tanx)/(x^2sinx)# as x approaches 0?

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Answer 1 · 2017-03-02T22:31:06

Simplify the expression as follows

#(sinx-tanx)/(x^2sinx)=(sinx(1-1/cosx))/(x^2*sinx)= (cosx-1)/(x^2*cosx)#

Multiply and divide with #1+cosx# you get

#[(cosx-1)*(1+cosx)]/[(x^2*cosx)*(1+cosx)]#

#[-(1-cos^2x)]/(x^2*(cos^2x+cosx))#

#[-sin^2x]/[x^2*(cos^2x+cosx)]#

#-(sinx/x)^2 *1/(cos^2x+cosx)#

Finally the limit is

#lim_(x->0) [ -(sinx/x)^2 *1/(cos^2x+cosx)]=#

#[-(lim_(x->0) sinx/x)^2)*1/(lim_(x->0)[cos^2x+cosx])]=#

#[-(1)*1/(1+1)]=-1/2#

Finally

#lim_(x->0) (sinx-tanx)/(x^2sinx)=-1/2#