Limits, Continuity and Differentiability

Concept:Analyze continuity and differentiability of $g(x)=x f(x)$ based on properties of $f$.Explanation:• Option A: False. $g$ is not always continuous at $x=0$.Counterexample: Let $f(x)=\begin{cases} \dfrac{1}{x}\sin\left(\dfrac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases}$.Then $g(x)=\begin{cases} \sin\left(\dfrac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases}$.$\lim\limits_{x \to 0} g(x)$ does not exist (oscillates between $-1$ and $1$), so $g$ is discontinuous at $0$.• Option B: True. If $f$ is continuous at $0$, then $\lim\limits_{h\to 0} f(h)=f(0)$ exists.$g'(0)=\lim\limits_{h\to 0}\dfrac{g(h)-g(0)}{h}=\lim\limits_{h\to 0}\dfrac{h f(h)}{h}=\lim\limits_{h\to 0} f(h)=f(0)$.Thus $g$ is differentiable at $0$.• Option C: False. If $g$ is differentiable at $0$, then $g'(0)=\lim\limits_{h\to 0} f(h)$ exists.But we only know the limit exists; $f(0)$ may not equal this limit, so $f$ need not be continuous at $0$.• Option D: True. Differentiability of $g$ at $0$ directly gives existence of $\lim\limits_{x \to 0} f(x)$ (equal to $g'(0)$).Hence the limit exists.Answer:Statements B and D are TRUE.