How do u find the limit of x^a ln x, a>0, when x tends to 0+?

Question

by using l'hopital rule

Answer 1

I assume that the function is (x^a) (ln x). Treat it as (ln x) / (x^(-a)). Then it is infinity by infinity form and l'hospital's rule is applicable. Differentiating the numerator as well as the denominator we get (-1/a)x^a. This goes to zero.