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...

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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**