Continuity of Functions: Quick Revision

A brief note on Continuity of Functions. Read it before a test. Try each question at the end before you open its answer.

Continuity at a point

• The test: $f$ is continuous at $a$ when $\displaystyle\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.
• Removable break: the limit exists, but $f(a)$ is missing or different. Redefine $f(a)$ to repair it.
• Jump: both one-sided limits exist but differ. No value at $a$ repairs it.
• Infinite break: a one-sided limit is not finite.

Intervals and piecewise functions

• Closed interval: continuous inside, right-continuous at $a$ and left-continuous at $b$.
• Algebra: sums, products, quotients (nonzero denominator) and compositions of continuous functions are continuous.
• Standard functions: polynomials, $|x|$, $\sin x$, $\cos x$ and $e^x$ are continuous everywhere. $[x]$ jumps at every integer.
• Piecewise functions: check each joining point. The left-hand limit, the right-hand limit and the value must be equal.

The Intermediate Value Theorem

• Statement: if $f$ is continuous on $[a, b]$ and $N$ lies between $f(a)$ and $f(b)$, then $f(c) = N$ for some $c$ in $(a, b)$.
• Roots: a sign change of a continuous function traps a root. No sign change proves nothing.
• Bisection: each halving halves the interval, leaving a length of $\dfrac{b - a}{2^n}$ after $n$ steps.

A worked example

Let $f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \ne 2$, and $f(2) = k$. Find $k$ so that $f$ is continuous at $x = 2$.

1. Simplify. For $x \ne 2$, $f(x) = x + 2$.
2. The limit. As $x \to 2$ from either side, $f(x) \to 4$.
3. The answer. Continuity needs $f(2) = 4$, so $k = 4$.

Practice with new numbers

A program to try

The program estimates the two one-sided limits of the function of the worked example and then shows a sign change of a cubic.

Mistakes to avoid

• Checking one side only. Both one-sided limits are needed.
• Forgetting the value. The limit and $f(a)$ must be equal.
• Wrong piece at a joining point. Use the left formula for the left-hand limit and the right formula for the right-hand limit.
• No sign change means no root. It does not. Two roots can sit between two positive values.

Check yourself

1. Find the two one-sided limits of $\dfrac{|x - 1|}{x - 1}$ at $x = 1$.

2. Where is $\dfrac{1}{x^2 - 25}$ discontinuous?

3. A continuous function has $f(0) = -3$ and $f(2) = 4$. What does the Intermediate Value Theorem say about $f(x) = 0$?

4. After how many halvings is an interval of length $1$ shorter than $0.1$?

5. Is $[x]$ continuous at $x = 2.5$?

Sources

• OpenStax, Calculus Volume 1, Chapter 2, section on continuity.
• NCERT, Mathematics Part I for Class XII, Chapter 5, section on continuity.