IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - find the limit

joyfrancis

The answer is inf.

Solution:

sinx/x when x tends to 0 = 1 {tending to}

sinx/x-sinx tends to infinity when x tends to 0

apply lh

=>cosx/1-cosx

put x=0

limit =inf.

.: we have 1^inf form

let the expression be h(x)

h(x)=e^(sinx/x)(2sinx-x/x-sinx)

=e^(2sin^2x-xsinx/x^2-xsinx)

We just have to find

limx-->0 2sin^2x-xsinx/x^2-xsinx

it is easy now,

since it is 0/0 form apply lh

it becomes =>(2sin2x-xcosx-sinx)/(2x-xcosx-sinx)

again this is 0/0 so again differentiate

you willl get

(4cos2x+xsinx-2cosx)/(2+xsinx-2cosx)

Put x=0

4+0-2/2+0-2

=2/0

=inf

We thus have the answer as e^inf=inf