Lim x-->0+ x^sin(x)?

Question

How would I solve that? I tried making it : ln y = sin(x)ln x but I'm not sure where to go from there.

Answer 1

Well that's quite good, now you have:

[x→0+]lim sin x ln x, which gives you the indeterminate form 0*-∞. So rewrite this as:

[x→0+]lim ln x/(1/sin x)

Which now gives you the indeterminate -∞/∞, so you can employ L'hopital's rule:

[x→0+]lim (1/x)/(-csc x cot x)

Algebraic manipulation yields:

[x→0+]lim (-sin x)/x tan x

Which beomes the determinate form:

-1*0 (since [x→0+]lim (-sin x)/x = -1 and [x→0+]lim tan x = 0)

Which is simply 0. So ln y=0, y=1. And we are done.

Edit: corrected a sign error.

Edit: corrected another sign error which I missed when correcting the first sign error. Also changed 0 to 0+ to improve accuracy.