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



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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.

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