Let ℝ denote the set of all real numbers. Let f : ℝ arrow ℝ be an...

Given $g(x) = x f(x)$ for all $x \in \mathbb{R}$.

For option A:
Let $f(x) = \dfrac{1}{x}$ for $x \neq 0$ and $f(0) = 0$.
Then $g(x) = 1$ for $x \neq 0$ and $g(0) = 0$.
$\lim_{x \to 0} g(x) = 1 \neq g(0)$.
Thus, $g$ is not necessarily continuous at $x = 0$. Option A is false.

For option B:
The derivative of $g$ at $x = 0$ is given by:
$g'(0) = \lim_{x \to 0} \dfrac{g(x) - g(0)}{x - 0} = \lim_{x \to 0} \dfrac{x f(x) - 0}{x} = \lim_{x \to 0} f(x)$.
If $f$ is continuous at $x = 0$, then $\lim_{x \to 0} f(x) = f(0)$.
Since $f(0)$ is a finite real number, $g'(0)$ exists and $g$ is differentiable at $x = 0$. Option B is true.

For option C:
Let $f(x) = 0$ for $x \neq 0$ and $f(0) = 1$.
Then $g(x) = 0$ for all $x \in \mathbb{R}$.
Here, $g$ is differentiable at $x = 0$ with $g'(0) = 0$.
However, $\lim_{x \to 0} f(x) = 0 \neq f(0)$, so $f$ is not continuous at $x = 0$. Option C is false.

For option D:
If $g$ is differentiable at $x = 0$, then $g'(0)$ exists.
Since $g'(0) = \lim_{x \to 0} \dfrac{g(x) - g(0)}{x} = \lim_{x \to 0} f(x)$, the limit $\lim_{x \to 0} f(x)$ must exist. Option D is true.

Answer: If $f$ is continuous at $x = 0$, then $g$ is differentiable at $x = 0$; If $g$ is differentiable at $x = 0$, then $\lim_{x \to 0} f(x)$ exists