Question

For a function \(f: \mathbb{R}\to\mathbb{R}\), consider the following statements:
(I) If \(\lim_{h\to0}\{f(x+h)-f(x-h)\} = 0\), then f is continuous on \(\mathbb{R}\).
(II) If \(\lim_{h\to0}\dfrac{f(x+h)-f(x-h)}{2h} = 0\), then f is differentiable on \(\mathbb{R}\).
Choose the correct answer.

• A. Both statements (I) and (II) are false
• B. Both statements (I) and (II) are true
• C. Only statement (I) is true
• D. Only statement (II) is true

Hint:

Test both hypotheses on a single point discontinuity like f(0)=1, f(x)=0 elsewhere.

Answer 1

The Correct Option is A

Step 1: Build a candidate counterexample. Define \(f(x) = 1\) if \(x = 0\) and \(f(x) = 0\) for \(x \ne 0\). This function has a jump only at the single point \(x=0\) and is otherwise identically zero.

Step 2: Check the hypothesis of statement (I) at every point. For \(x \ne 0\) and \(h\) small enough that \(x+h \ne 0\) and \(x-h \ne 0\), \(f(x+h) - f(x-h) = 0 - 0 = 0\). At \(x = 0\), for any \(h \ne 0\), \(f(h) - f(-h) = 0 - 0 = 0\) since \(h \ne 0\) and \(-h \ne 0\). So \(\lim_{h\to0}\{f(x+h)-f(x-h)\} = 0\) holds for every \(x \in \mathbb{R}\), meaning \(f\) satisfies the hypothesis of statement (I) everywhere.

Step 3: Check the conclusion of (I). But \(f\) is not continuous at \(x=0\), since \(f(0) = 1\) while \(\lim_{x\to0} f(x) = 0 \ne f(0)\). The hypothesis holds while the conclusion fails: statement (I) is false.

Step 4: Check the hypothesis of statement (II) with the same function. For \(x \ne 0\), as in Step 2, \(f(x+h)-f(x-h) = 0\) for small \(h\), so the quotient \(\dfrac{f(x+h)-f(x-h)}{2h} = 0\) for all such \(h\), giving limit \(0\). At \(x=0\), for any \(h \ne 0\), \(\dfrac{f(h)-f(-h)}{2h} = \dfrac{0-0}{2h} = 0\). So the symmetric-derivative hypothesis of statement (II) also holds at every \(x \in \mathbb{R}\).

Step 5: Check the conclusion of (II). Since \(f\) is not even continuous at \(x=0\), it cannot be differentiable there. The hypothesis holds while the conclusion fails: statement (II) is also false.

Step 6: Conclude. The same function \(f\) disproves both implications, so both statements are false. \[\boxed{\text{Both statements (I) and (II) are false}}\]