Find the limit of
lim (sin x - x)/x^3
As x approaches 0?
By L'Hopital's rule, we need to find the lim as x approaches 0 of (cos x - 1)/(3x^2). (Note we take the derivative of the numerator and denominator.
We can repeat this again to find that this limit is equivalent to lim as x approaches 0 of (-sin x)/(6x) which is equal to lim as x approaches 0 of (-cosx)/6 = -1/6.
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