Find the limit as x -> 0+ of (sin x)(ln x)
The Attempt at a Solution
I rewrote this as (sin x) / (1/ln x), then using L'Hopital it becomes:
(cos x) / [(-ln x) / (ln x)2] = [(ln x)2 cos x] / (-ln x)
So I get limit as x -> 0+ of [(ln x)2 cos x] / (-ln x)
Which isn't any better. I'm not sure if I can even use L'Hopital here because it requires that substitution gives the indeterminate state, but I don't know if 0+/0+ really counts. Any thoughts?