Question

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

Answer 1

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))`

Explanation:

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