Solving limit with hopital

Question

lim as x tends to 0+ (x^1/2)(lnx)

Answer 1

Limit as x→0+ of x^1/2 lnx = ?

 

Rewrite as

 

lnx / x^-1/2

 

Apply L'Hopital's rule.

 

lnx / x^-1/2 → (1/x) / [(-1/2)x^-3/2] = -2x^3/2 / x = -2x^1/2 → 0 as x→0+

 

The limit is 0.