1.4 · Continuity, discontinuities, and the Intermediate Value Theorem

The three-part continuity test

A function is continuous at x = c if f(c) is defined, the finite two-sided limit $\displaystyle \lim_{x\to c}f(x)$ exists, and that limit equals f(c). Check the conditions in that order. A graph with a hole can fail the first condition; a jump fails the second; a misplaced filled dot fails the third.

Polynomials are continuous on the real line. Rational functions are continuous wherever their denominator is nonzero. Radicals, logarithms, and trigonometric functions require attention to their domains. Sums, products, and compositions of continuous functions remain continuous wherever the operations are defined.

$$\lim_{x\to c}f(x)=f(c)$$

Classify the break before trying to repair it

A removable discontinuity occurs when a finite two-sided limit exists but the function is undefined there or has the wrong value. Setting f(c) equal to that limit repairs the break. A jump has finite but unequal one-sided limits. An infinite discontinuity involves unbounded behavior near the point.

Changing a single point value cannot fix a jump or a vertical asymptote. For $\displaystyle \frac{x^2-1}{x-1}$, the hole at 1 is removable because the limit is 2. By contrast, a function with left limit 2 and right limit 5 has a jump, regardless of its value at the joining point.

Make piecewise functions join

At a joining input c, use the left branch to find the left-hand limit, the right branch to find the right-hand limit, and the branch that includes equality to find f(c). Choose any parameters so all three values agree.

For $\displaystyle f(x)=ax+1$ when x < 2 and $\displaystyle f(x)=x^2-1$ when x ≥ 2, the right limit and f(2) are 3. The left limit is $\displaystyle 2a+1$. Setting $\displaystyle 2a+1=3$ gives a = 1, making the function continuous at the join.

$$\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)$$

Continuity on intervals and at endpoints

A function is continuous on an open interval if it is continuous at every point inside it. For continuity on [a, b], also require $\displaystyle \lim_{x\to a^+}f(x)=f(a)$ and $\displaystyle \lim_{x\to b^-}f(x)=f(b)$. At endpoints you approach from inside the interval.

For example, $\displaystyle \sqrt{x+1}$ is continuous on [−1, 3]: it is continuous in the interior, right-continuous at −1, and left-continuous at 3. For $\displaystyle \frac{1}{x-2}$, the maximal open intervals of continuity are (−∞, 2) and (2, ∞). A denominator restriction cannot be ignored just because the requested interval crosses it.

The Intermediate Value Theorem guarantees existence

If f is continuous on [a, b] and N lies strictly between f(a) and f(b), then some c in (a, b) satisfies f(c) = N. The function cannot move continuously between two heights without taking every intermediate height.

For a root, use N = 0 and show opposite signs at the endpoints. State continuity on the whole closed interval, evaluate the endpoints, show that 0 lies between those values, and conclude that at least one root exists in the open interval. IVT does not give an exact root or establish uniqueness.

$$f(c)=N,\quad c\in(a,b)$$

Know what IVT cannot conclude

Opposite signs alone are not enough: $\displaystyle f(x)=\frac{1}{x}$ has values −1 and 1 at −1 and 1, but it is not continuous on [−1, 1] and never equals 0. The continuity hypothesis prevents this false inference.

If endpoint values have the same sign, IVT gives no root guarantee from those endpoints; a root may still exist. For instance, x² on [−1, 1] has positive endpoint values but has a zero at 0. Distinguish “not guaranteed by these data” from “impossible.”