Limit using hopital

Question

lim x tends to 0+ (sinx)(lnt)

Answer 1

I assume you meant 

 

Limit as x→0+ of (sinx)(lnx)

 

Rewrite (sinx)(lnx) as lnx / (1/sinx) and apply L'Hopital's rule.

 

lnx / (1/sinx) → (1/x) / [(-1/sin²x)cosx] = -sin²x / (x cosx) 

 

The resulting form is still indeterminate, so apply the rule again and get

 

-sin²x / (x cosx) → -2sinx cosx / (cosx - x sinx) → -2(0)(1) / (1 -0*0) = 0 as x→0

 

The limit is zero.