What is $lim_ (x->0) (xcosx)/(sin3x)$ ?

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Answer 1 · 2016-11-22T08:28:12

Since the outcome implies " $0/0$ which is indeterminate... "

We can apply L'Hopital's Rule:

IF $lim_(x->a)f(x)/g(x) = 0/0 or oo/oo$

Then $lim_(x->a)f(x)/g(x)=^(L'H)lim_(x->a)(f'(x))/(g'(x))$

$lim_(x->0)(xcosx)/sin(3x) =^(L'H)lim_(x->0)(cosx-xsinx)/(3cos(3x))=1/3$

What is lim_ (x->0) (xcosx)/(sin3x)lim_ (x->0) (xcosx)/(sin3x) ?