Question

lim x->0 (e^x-e^sinx)/(x-sinx) solve by l'hospital rule

Answer 1

→0

ex

−

e

−x

−2x

x−

sinx

=

e0

−

e

−0

−2

(0)

0−

sin0

=

1−1−0

0−0

=00

This is an indeterminate type so use l'Hopital's Rule which is the limit of the quotient of the derivative of the top and the derivative of the bottom as x goes to 0.

lim

x→0

ex

+

e

−x

−2

1−

cosx

=

e0

+

e

−0

−2

1−

cos0

=

1+1−2

1−1

Answer 2

Ans. is 1 ... solution is in the pic....
we have to differentiate it three times ....