Continuity & Differentiability of Composite Functions
DAILY_CHALLENGE
Grade None
Question:
Let $\mathbb{R}$ denote the set of all real numbers. Let $f:\mathbb{R}\to\mathbb{R}$ be an arbitrary function and let $g:\mathbb{R}\to\mathbb{R}$ be the function defined by
$$g(x) = xf(x), \quad \text{for all } x \in \mathbb{R}.$$
Then which of the following statements is (are) TRUE?
The function $g$ is always continuous at $x = 0$
If $f$ is continuous at $x=0$, then $g$ is differentiable at $x=0$
If $g$ is differentiable at $x=0$, then $f$ is continuous at $x=0$
If $g$ is differentiable at $x=0$, then $\displaystyle\lim_{x\to 0} f(x)$ exists
Step-by-Step Solution
Key Concept: Differentiability of $g(x)=xf(x)$ at $x=0$ forces $\lim_{x\to0}f(x)$ to exist (equal to $g'(0)$), but does NOT require $f(0)$ to match that limit — continuity of $f$ is a strictly stronger condition.
**Step 1: Check (A): is $g$ always continuous at $0$?**
Counterexample: $f(x) = 1/x$ for $x\ne0$, $f(0)=0$. Then $g(x)=1$ for $x\ne0$ and $g(0)=0$, so $g$ is discontinuous at $0$. (A) is FALSE.
**Step 2: Check (B): $f$ continuous $\Rightarrow$ $g$ differentiable at $0$**
$g'(0) = \lim_{x\to0}\dfrac{g(x)-g(0)}{x} = \lim_{x\to0}\dfrac{xf(x)}{x} = \lim_{x\to0}f(x) = f(0)$ (by continuity of $f$). So $g'(0)$ exists. (B) is TRUE.
**Step 3: Check (C): $g$ differentiable $\Rightarrow$ $f$ continuous at $0$?**
$g'(0)$ existing forces $\lim_{x\to0}f(x) = g'(0)$, but $f(0)$ could be any value $\ne g'(0)$. Counterexample: $f(x)=1$ for $x\ne0$, $f(0)=0$. Then $g(x)=x$ for $x\ne0$, $g(0)=0$, so $g'(0)=1$ exists, but $f$ is discontinuous. (C) is FALSE.
**Step 4: Check (D): $g$ differentiable $\Rightarrow$ $\lim_{x\to0}f(x)$ exists**
$g'(0) = \lim_{x\to0}\dfrac{xf(x)}{x} = \lim_{x\to0}f(x)$. For $g'(0)$ to exist, this limit must exist and equal $g'(0)$. (D) is TRUE.
Correct Answer: B, D