# What is limit [(1/sinx) - (1/x)] as x approaches 0+ ? Thank you for answering :)

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We'll verify first if calculating the limit, we'll get an indeterminate form:

lim [(1/sinx) - (1/x)] = lim [(x - sin x)/x*sin x]

lim [(x - sin x)/x*sin x] = (0 - sin 0)/(0*sin0) = 0/0

Since we've get the indeterminate form "0/0" type, we'll use L'Hospital rule to determine the limit:

lim f/g = lim f'/g'

Let f = x - sin x => f' = 1 - cos x

Let g = x*sin x => g' = sin x + x*cos x

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

We'll substitute x by the value of accumulation point x = 0:

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

(1-cos 0)/(sin 0 + 0*cos 0) = (1-1)/(0+0) = 0/0

Since we've get an indetermination form again, we'll apply L'Hospital rule, once more time;

lim (1-cos x)/(sin x + x*cos x) = lim (1-cos x)'/(sin x + x*cos x)'

lim (1-cos x)'/(sin x + x*cos x)' = lim sin x/(cos x + cosx - x*sin x)

We'll substitute x by the value of accumulation point x = 0:

lim sin x/(cos x + cosx - x*sin x) = sin 0/(cos 0 + cos0 - 0*sin 0)

sin 0/(cos 0 + cos0 - 0*sin 0) = 0/(1+1-0) = 0/2 = 0

**The limit of the given function, if x approaches to zero, is lim [(1/sinx) - (1/x)] = 0.**