Answer
See explanations.
Work Step by Step
Step 1. Identify the given conditions: $f(x_0)=0$, $f(x)$ is differentiable at $x_0$ and $g(x)$ is continuous at $x_0$.
Step 2. Letting $h(x)=f(x)g(x)$, we have $h'(x_0)=\lim_{x\to x_0}\frac{h(x)-h(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{f(x)g(x)}{x-x_0}=\lim_{x\to x_0}\frac{f(x)}{x-x_0}\times\lim_{x\to x_0}g(x)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\times\lim_{x\to x_0}g(x)=f'(x_0)g(x_0)$
Step 3. Using the conditions given in step 1, both $f'(x_0)$ and $g(x_0)$ exist; thus $h'(x_0)$ also exists.
