Continuity and Differentiability

Core ≈ 55 min Chapter 5 8 exercises
Why this mattersContinuity is what lets calculus trust the world between the points we know, and the derivative is what measures change at a single instant. Every later chapter — rules, applications, integrals — is built on these two ideas.

1 · Continuity at a point — the three conditions

f is continuous at x = a when all three hold:

  1. f(a) is defined,
  2. lim x→a f(x) exists,
  3. lim x→a f(x) = f(a).

If any fails, f has a discontinuity at a: removable (a hole — factor cancels), jump (one-sided limits disagree), or infinite (vertical asymptote). Polynomials are continuous everywhere; rational functions are continuous everywhere except where the denominator vanishes.

2 · Continuity on an interval — and the IVT

f is continuous on [a, b] when it is continuous on (a, b) and continuous from the right at a and from the left at b. Continuous-on-an-interval is the hypothesis that unlocks the two big existence theorems:

Intermediate Value Theorem (IVT): if f is continuous on [a, b] and N lies between f(a) and f(b), some c in (a, b) has f(c) = N. Sign change between endpoints ⇒ a root inside.

Worked example Example 1 — Show x³ + x − 1 = 0 has a root in (0, 1).
Step 1: f(x) = x³ + x − 1 is a polynomial, hence continuous everywhere.
Step 2: f(0) = −1 (negative); f(1) = 1 + 1 − 1 = 1 (positive).
Step 3: 0 lies between −1 and 1, so by the IVT some c in (0, 1) satisfies f(c) = 0.
Notice: the IVT guarantees a root exists without finding it. That is why it is the standard tool for proving existence (and for bisection algorithms on computers).

3 · Extreme Value Theorem — max and min exist

Extreme Value Theorem (EVT): a continuous function on a closed, bounded interval [a, b] attains an absolute maximum and an absolute minimum there. Both hypotheses matter — on an open interval or with a blow-up, the max/min can escape to the boundary or to infinity. EVT is why optimization problems on [a, b] always have an answer, and why closed intervals are non-negotiable in the fine print.

4 · From average velocity to instantaneous velocity

If s(t) is position, average velocity over [t, t + h] is Δs/Δt = (s(t + h) − s(t))/h — displacement over time. Shrink h toward 0 and the averages settle onto the instantaneous velocity: v(t) = lim h→0 (s(t + h) − s(t))/h. Displacement is change in position (can be negative); total distance traveled adds the absolute values. Speed is |v(t)|, always nonnegative.

5 · Tangent lines and the derivative as a limit

The tangent line at (a, f(a)) is the line that best matches f's direction there. Its slope is the derivative:

f′(a) = lim h→0 [f(a + h) − f(a)] / h

Equivalently, f′(a) = lim x→a [f(x) − f(a)] / (x − a). The derivative function f′(x) assigns to every x where the limit exists the slope of the tangent there. A linear function f(x) = mx + b has f′(x) = m — constant everywhere, the anchor example for everything that follows.

Worked example Example 2 — Differentiate f(x) = x² from the definition.
Step 1: set up the quotient: [(x + h)² − x²] / h.
Step 2: expand: (x² + 2xh + h² − x²)/h = (2xh + h²)/h = 2x + h (valid for h ≠ 0).
Step 3: let h → 0: f′(x) = 2x.
So the tangent to y = x² at x = 3 has slope 6 — the parabola's slope grows without bound as you slide right, and it is exactly 0 at the vertex.

The second derivative f″(x) = (f′)′(x) measures the rate of change of the rate of change — acceleration when s is position, or the bending (concavity) of a graph. f″ > 0 means curving upward; f″ < 0 curving downward.

6 · When derivatives fail to exist — and the one-way street

Differentiability is stronger than continuity. Where f′ fails to exist:

  • Corners/cusps — |x| at 0: left slope −1, right slope +1, no single tangent.
  • Vertical tangents — x^(1/3) at 0: the tangent is vertical, slope unbounded.
  • Discontinuities — jumps and holes never have tangent slopes.

Differentiability implies continuity: if f′(a) exists, f must be continuous at a (proof sketch: f(x) − f(a) = [(f(x) − f(a))/(x − a)]·(x − a) → f′(a)·0 = 0). The converse is false — |x| is continuous at 0 but not differentiable there.

Watch outDo not read f′(x) = 0 as "f is flat everywhere" or "f has a maximum" — it only says the tangent is horizontal at that point. Peaks, troughs, plateaus and inflection-level terraces all pass through horizontal tangents.
Chapter formula sheet f continuous at a ⇔ f(a) defined · lim x→a f(x) exists · lim x→a f(x) = f(a)
IVT: f continuous on [a,b], N between f(a) and f(b) ⇒ ∃ c with f(c) = N
EVT: f continuous on [a,b] ⇒ f attains absolute max and min on [a,b]
f′(a) = lim h→0 [f(a + h) − f(a)] / h = lim x→a [f(x) − f(a)] / (x − a)
f linear ⇒ f′ = m · v = s′ · a = v′ = s″ · differentiable ⇒ continuous (not conversely)

Common mistakes

  • Checking only two of the three continuity conditions — defined, limit exists, and equal are all required.
  • Applying the IVT without verifying continuity, or reading it as "there is exactly one root".
  • Assuming differentiability from continuity — |x| at 0 is the standing counterexample.
  • Computing average velocity and calling it instantaneous; shrink the interval first, or take the limit.
  • Sign errors in the difference quotient: it is f(a + h) − f(a) over h, not f(a) − f(a + h).
  • Thinking f″ > 0 means the function is increasing — it means concave up; f′ is what measures increase.

✎ Practice — marked instantly

f is continuous at x = a when:
  • A. f(a) is defined
  • B. lim x→a f(x) exists
  • C. lim x→a f(x) = f(a)
  • D. all three of the above

Multiple choice · interactive marking is enabled when JavaScript loads.

Let f(x) = (x² − 1)/(x − 1) for x ≠ 1, and f(1) = 2. Is f continuous at x = 1? Answer yes or no.

Short answer · interactive marking is enabled when JavaScript loads.

The IVT allows you to conclude that x³ + x − 1 = 0 has a root in (0, 1) because:
  • A. f(0) and f(1) have opposite signs and f is continuous
  • B. f is a polynomial
  • C. the function is increasing
  • D. the derivative never vanishes

Multiple choice · interactive marking is enabled when JavaScript loads.

A particle has position s(t) = t² + 1 (meters, seconds). What is its average velocity over [1, 3] in m/s?

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Using f(x) = x² and the limit definition, find f′(3).

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Which statement is TRUE?
  • A. Continuous ⇒ differentiable
  • B. Differentiable ⇒ continuous
  • C. Both are equivalent
  • D. Neither implies the other

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Compute the derivative of f(x) = 5x + 7 (linear function). Enter f′(x).

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At a corner point such as f(x) = |x| at x = 0, which condition for differentiability fails?
  • A. f is not continuous there
  • B. The left- and right-hand difference quotients disagree
  • C. f(0) is undefined
  • D. The limit of f(x) as x → 0 does not exist

Multiple choice · interactive marking is enabled when JavaScript loads.