What is limit [(1/sinx) - (1/x)] as x approaches 0

We have to find the value of lim x -->0 [(1/sinx) -
(1/x)]

if we substitute x = 0 we get an indeterminate form
of inf. - inf.

lim x -->0 [(1/sinx) -
(1/x)]

=> lim x -->0 [(x/x*sinx) - (sin
x/x*sin x)]

=> lim x -->0 [(x - sin
x)/x*sinx)]

substitute x = 0, we get the indeterminate form
0/0. This allows us to use l"Hopital's rule and substitute the numerator and the
denominator with their derivatives.

=> lim x
-->0 [(1 - cos x)/(x*cos x + sin x)]

substitute x =
0

=> (1- 1)/(0-0)

again
the indeterminate form 0/0, use l'Hopital's rule
again

=> lim x -->0 [(sin x)/(cos x - x*sin x
+ cos x)]

=> lim x -->0 [(sin x)/(2*cos x -
x*sin x)]

substitute x =
0

=> 0/ 2

=>
0

The value of lim x -->0 [(1/sinx) -
(1/x)] = 0