Question 7

Analyze each statement:

(A) $g(x) = x f(x)$. If $f(x) = \frac{1}{x^2}$ for $x \neq 0$, then $\lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{1}{x}$, which does not exist. So $g$ is not always continuous. (False)

(B) By definition, $g'(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0} \frac{h f(h) - 0 \cdot f(0)}{h} = \lim_{h \to 0} f(h)$. If $f$ is continuous at $x=0$, $\lim_{h \to 0} f(h) = f(0)$, so $g'(0)$ exists and equals $f(0)$. (True)

(C) If $g$ is differentiable at $x=0$, then $\lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0} f(h)$ must exist. However, for $f$ to be continuous at $x=0$, we need $\lim_{h \to 0} f(h) = f(0)$. $g$ being differentiable at $0$ does not depend on the specific value of $f(0)$.

For example, if $f(x)=1$ for $x \neq 0$ and $f(0)=0$, then $g(x)=x$ for all $x$, which is differentiable, but $f$ is discontinuous at $0$. (False)

(D) As shown in (B), the existence of $g'(0)$ is equivalent to the existence of $\lim_{x \to 0} f(x)$. (True)