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Question

limx2sin(ex21)log(x1)

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Solution

We have, limx2sin(ex21)log(x1)

At x=2, the value of the given expression

sin(e221)log(21)=sin0log1=00form

Applying L'Hospital Rule

limx2ddx(sin(ex21))ddxlog(x1)

limx2cos(e(x2)1).ddx{e(x2)1}1x1ddx(x1)

limx2cos(e(x2)1)ex21(x1)1

limx2(x1)cos(e(x2)1)ex2

Putting x=2

(2-1) cos(e221)×e(22)

=1×cos 0×e0

=1×1×1=1


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