The limit as x approaches zero of (sinx - xcosx)/x^3

a2linetagadaW 2021-03-04 Answered
The limit as x approaches zero of sinxxcosxx3
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rogreenhoxa8
Answered 2021-03-05 Author has 109 answers

Let f(x)=g(x)h(x),where g(x)=sinxx.cosx,andh(x)=x3.
The limit is x0.
Since g(0) and h(0) both =0,then use l'Hopital's rule, limf(x)[x>0]=lim[x>0]g(x)/h(x)
g(x)=cosxcosx+x.sinx=x.sinx
h(x)=3x2
Since g'(0) and h'(0) both = 0,then again use l'Hopital's rule, i.e. limf(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. limf(x)[x>0]=lim[x>0]g(x)/h(x)
g(x)=cosx+cosxx.sinx=2cosxx.sinx
h(x)=6
Now, g(0)=2, and h(0)=6.
So, limf(x)[x>0]=lim[x>0]g(x)/h(x)=lim[x>0]2/6=1/3

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