Let f : R → R be defined by f(x) = x·sin(1/x) if x > 0, and f(x) = 0 if x ≤ 0.…

Concept: A function is continuous at a point when its left-hand limit, right-hand limit, and function value there all agree; the Squeeze (Sandwich) Theorem gives a limit of 0 whenever a function is trapped between two bounds that both tend to 0. Differentiability at a point additionally requires the left-hand and right-hand difference quotients, lim as h→0 of [f(a+h) − f(a)]/h, to both exist and match — and differentiability always forces continuity, though continuity never forces differentiability.

Application: Test f at x = 0 using these ideas in turn.

1. For x > 0, f(x) = x·sin(1/x). Since |sin(1/x)| ≤ 1 for every x ≠ 0, |f(x)| = |x|·|sin(1/x)| ≤ |x|, and |x| → 0 as x → 0⁺. By the squeeze theorem, the right-hand limit of f at 0 is 0.

2. For x ≤ 0, f(x) = 0 identically, so the left-hand limit of f at 0 is 0, and f(0) = 0 by definition.

3. The right-hand limit, left-hand limit, and f(0) are all equal to 0, so f is continuous at x = 0.

4. For the derivative, take h < 0: [f(h) − f(0)]/h = (0 − 0)/h = 0, so the left-hand derivative f′(0⁻) = 0.

5. Take h > 0: [f(h) − f(0)]/h = [h·sin(1/h) − 0]/h = sin(1/h).

6. As h → 0⁺, sin(1/h) keeps oscillating between −1 and 1 and never settles on one value, so the limit defining the right-hand derivative f′(0⁺) does not exist.

Cross-check: Approach 0⁺ along two explicit sequences to confirm the oscillation directly — hₙ = 1/(nπ) gives sin(1/hₙ) = sin(nπ) = 0 for every n, while hₙ′ = 1/((2n + 0.5)π) gives sin(1/hₙ′) = sin((2n + 0.5)π) = 1 for every n. Both sequences of h-values tend to 0, yet the corresponding difference quotients tend to two different numbers (0 and 1), so no single right-hand derivative can exist — an independent confirmation of the previous step.

Result: f is continuous at x = 0 (all three values match) but not differentiable there (the right-hand derivative fails to exist).