Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 44

You need to evaluate the limit such that:
lim_(x->0^+) sin x*ln x = 0*(-oo)
You need to use the special limit lim_(x->0^+) (sin x)/x = 1 , such that:
lim_(x->0^+) ((sin x)/x)*x*ln x= 1*lim_(x->0^+)x*ln x = 0*(-oo)
You need to convert the product into a quotient for the l'Hospital's rule to be applied:
lim_(x->0^+) (ln x)/(1/x) = lim_(x->0^+) ((ln x)')/((1/x)')
lim_(x->0^+) (1/x)/(-1/x^2) = lim_(x->0^+) (-x) = 0
Hence, evaluating the limit of the function, using special limit and l'Hospital's rule yields lim_(x->0^+) sin x*ln x = 0.