Solve for lim x-->0 [(x – sin x) / x^3]

Here if replace x with 0 we get the result as 0/0 which is
indeterminate. Therefore we can use L’ Hopital’s Rule which states that if an expression
of the form lim x-->0 [f(x)/ g(x)] gives the indeterminate form 0/0, the limit
can be found as [f’(x)/g’(x)] for x=0.

Now for the given
expression [(x – sin x) / x^3], f(x) = x – sin x and g(x) =
x^3

f’(x) = (1- cos x)

g’(x) =
3x^2

Now we have [f’(x)/g’(x)] for x=0 as (1- cos x) / 3x^2
= 0/0 again for x =0.

So we take the differential
again

f’’(x) = sin x

g’’(x) =
6x

If we determine sin x / 6x for x=0 we get 0/0
again

So we take the differential
again

f’’’(x) = cos x

g’’’(x)
= 6

Now cos x / 6 for x= 0 = 1/6, which is not
indeterminate.

Therefore the result is
1/6.