differentiability implies continuity

If a graph is smooth enough to have a well-defined slope at a point, it certainly cannot have a hole or a jump there — you cannot tilt a ruler against a gap. Having a derivative is a strong condition, and one of the first things it buys you, for free, is continuity at that point.

The theorem: if f is differentiable at a, then f is continuous at a. The proof is a one-line product of limits. For x near a but not equal to a, write f(x) - f(a) = ((f(x) - f(a)) / (x - a)) * (x - a). As x -> a the first factor tends to f'(a) (a finite number, by differentiability) and the second tends to 0, so the product tends to 0. Hence f(x) -> f(a), which is exactly continuity at a.

The implication runs in one direction only; its converse is false. Continuity does not imply differentiability: f(x) = |x| is continuous at 0 but has no derivative there, and worse examples (continuous everywhere, differentiable nowhere) exist. So differentiability is a sufficient condition for continuity but far from necessary, and the two notions must never be conflated.