Continuity and Differentiability Analysis of Product Function at Origin

To determine which of the given statements are true for the function $g(x) = x f(x)$, where $f: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function, let us analyze each option individually using the definitions of continuity and differentiability.

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Analysis of Option (A):

"The function $g$ is always continuous at $x = 0$"

Consider a counterexample: Let $f(x)$ be defined as $f(x) = \begin{cases} \frac{1}{x^2}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ Then $g(x) = x f(x)$ is given by $g(x) = \begin{cases} \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ Evaluating the limit of $g(x)$ as $x \to 0$: $\lim_{x \to 0^+} g(x) = \infty \quad \text{and} \quad \lim_{x \to 0^-} g(x) = -\infty$ Since $\lim_{x \to 0} g(x)$ does not exist, $g(x)$ is not continuous at $x = 0$.

Thus, Option (A) is FALSE.

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Analysis of Option (B):

"If $f$ is continuous at $x = 0$, then $g$ is differentiable at $x = 0$"

By definition, $g$ is differentiable at $x = 0$ if the following limit exists: $g'(0) = \lim_{h \to 0} \frac{g(0 + h) - g(0)}{h}$ We know $g(0) = 0 \cdot f(0) = 0$ and $g(h) = h f(h)$. Substituting these into the derivative definition: $g'(0) = \lim_{h \to 0} \frac{h f(h) - 0}{h} = \lim_{h \to 0} f(h)$ Since $f$ is continuous at $x = 0$, we have: $\lim_{h \to 0} f(h) = f(0)$ Because this limit exists and equals $f(0)$, $g$ is differentiable at $x = 0$ with $g'(0) = f(0)$.

Thus, Option (B) is TRUE.

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Analysis of Option (C):

"If $g$ is differentiable at $x = 0$, then $f$ is continuous at $x = 0$"

Consider a counterexample: Let $f(x)$ be defined as $f(x) = \begin{cases} 1, & x \neq 0 \\ 5, & x = 0 \end{cases}$ Then $g(x) = x f(x)$ becomes: $g(x) = x \quad \text{for all } x \in \mathbb{R} \quad (\text{since } g(0) = 0 \cdot 5 = 0)$ Here, $g(x) = x$ is a linear function and is differentiable everywhere, including at $x = 0$, with $g'(0) = 1$. However, for $f(x)$: $\lim_{x \to 0} f(x) = 1 \neq f(0) = 5$ Thus, $f$ is not continuous at $x = 0$.

Thus, Option (C) is FALSE.

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Analysis of Option (D):

"If $g$ is differentiable at $x = 0$, then $\lim_{x \rightarrow 0} f(x)$ exists"

If $g$ is differentiable at $x = 0$, then by definition of the derivative at $x = 0$: $g'(0) = \lim_{x \to 0} \frac{g(x) - g(0)}{x - 0}$ Using $g(0) = 0 \cdot f(0) = 0$ and $g(x) = x f(x)$: $g'(0) = \lim_{x \to 0} \frac{x f(x) - 0}{x} = \lim_{x \to 0} f(x)$ Since $g$ is differentiable at $x = 0$, $g'(0)$ is a well-defined finite real number. Consequently, $\lim_{x \to 0} f(x)$ exists and is equal to $g'(0)$.

Thus, Option (D) is TRUE.

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Conclusion:

The correct statements are B and D.