Q. lim x tends to pi/2

1+cos2x / (pi-2x)2

Given, limxπ21+cos2x(π2x)2=limxπ22cos2x4(π2x)2=12limxπ2cos2x(π2x)2=12limxπ2[sin(π2x)]2(π2x)2=12limy0(sinyy)2 where (π2x)=y=>y0.=12(limy0sinyy)2=12×12 (Since limy0sinyy=1 )=12.

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Let (pi -2x) =y then 2x= (pi-y) and lim x tends to pi /2 can be written as lim y--0

lim y---0 (1+cos(pi-y) )/y2

= lim y---0 (1-cos y )/y2 ------------As cos(pi-x) = -cosx--------------

= lim y--0 (-2sin2(y/2) /y2 ---------------As cosy = 1-2sin2(y/2)-------

= -1/2 lim y--0 [(sin(y/ 2 )/(y/2)] 2 =-1/2 ------------Multiplyin by 4 to numerator and denominator--------

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=lim y--0 [ "+" 2sin2(y/2) / y2 ] ---------------

Ans: 1/2

Anyways 10q vei mch..! :)

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