Question

Value of limx → 0⁡(1+Sin(x))Cosec(x)​

Answer 1

Answer:

Let the limit = L

L = lim x-> 0 ( 1 + sinx )^cosecx

logL = lim x-> 0 cosecx [ log( 1 + sinx ) ]

logL = lim x-> 0 [ log( 1 + sinx ) ] / sinx

This is a 0/0 form.

Hence we apply the L'hospital rule.

logL = lim x-> 0 [ 1/( 1 + sinx ) ] cosx/cosx

logL = 1

L = e.