Question

Question #57637

Answer 1

`lim_(x->0^+) sinx lnx =0`

Explanation:

This limit is in the indeterminate form `0*oo`. To apply L'Hospital's rule we have to transform it in a way that it is in the form `0/0` or `oo/oo`:

`lim_(x->0^+) sinx lnx = lim_(x->0^+) lnx/(1/sinx)`

we can now apply the rule and have:

`lim_(x->0^+) lnx/(1/sinx) = lim_(x->0^+) (d/dx lnx)/(d/dx 1/sinx) = lim_(x->0^+) (1/x)/(-cosx/sin^2x) = lim_(x->0^+) -sinx /x tanx = -1*0 = 0`

graph{sinxlnx [-10, 10, -5, 5]}