# `lim_(x->0^+) sin(x)ln(x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply,...

`lim_(x->0^+) sin(x)ln(x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Expert Answers

sciencesolve | Certified Educator

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.` **