Question

How do you find the limit of `ln(sinx)` as `x->0^-`?

Answer 1

The limit does not exist since `ln(sin x)` is not defined on `(-pi,0)`.

Explanation:

Since `sin(x)<0` on `(-pi,0)`, `ln(sin x)` is not defined on `(-pi,0)` as you can see in the graph below.

Hence, the limit does not exist.