Question: the limit as x approaches zero of (sinx - xcosx)/x^3 .
Let f(x) = g(x)/h(x), where g(x) = sinx - x.cosx, and h(x) = x^3.
The limit is x -> 0.
Since g(0) and h(0) both = 0,then use l'Hopital's rule, viz. Lim f(x) [x -> 0] = Lim [x -> 0] g'(x) / h'(x)
g'(x) = cosx - cosx + x.sinx = x.sinx
h'(x) = 3x^2
Since g'(0) and h'(0) both = 0,then again use l'Hopital's rule, i.e. Lim f(x) [x -> 0] = Lim [x -> 0] g''(x) / h''(x)
g''(x) = sinx + x.cosx
h''(x) = 6x
Since g''(0) and h''(0) both = 0,then again use l'Hopital's rule, i.e. Lim f(x) [x -> 0] = Lim [x -> 0] g'''(x) / h'''(x)
g'''(x) = cosx + cosx - x.sinx = 2cosx - x.sinx
h'''(x) = 6
Now, g''(0) = 2, and h'''(0) = 6.
So, Lim f(x) [x -> 0] = Lim [x -> 0] g'''(x) / h'''(x) = Lim [x -> 0] 2/6 = 1/3
Answer: the limit is 1/3