Identify discontinuities

A discontinuity is a point at which a function is undefined, its two-sided limit does not exist, or the limiting value differs from the function’s value; these can be identified from graphs, tables, and formulas such as piecewise and rational functions. The common cases—removable discontinuities (holes), jump discontinuities, and infinite discontinuities associated with vertical asymptotes—are distinguished using domain and one-sided or two-sided limits, without requiring formal epsilon–delta definitions or pathological examples.

Detailed Explanation: Identify discontinuities

A function is discontinuous at $x=a$ if at least one of these occurs:

$f(a)$ is undefined.
The two-sided limit $\displaystyle \lim_{x\to a}f(x)$ does not exist.
The limit exists, but it is not equal to $f(a)$ .

To identify discontinuities, check:

where a denominator is zero,
where a piecewise formula changes,
where there is a missing point or specially defined value.

Worked example

Consider

f(x)= \begin{cases} \dfrac{x^2-1}{x-1}, & x<2,\ x\ne 1,\\[6pt] 5, & x=1,\\ 4, & x=2,\\[4pt] \dfrac{1}{x-3}, & x>2. \end{cases}

We will check the possible problem points $x=1$ , $x=2$ , and $x=3$ .

Step 1: Check $x=1$

For values near $1$ , use

\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1, \qquad x\ne 1.

Therefore,

\lim_{x\to 1}f(x) = \lim_{x\to 1}(x+1) = 2.

However, the function is defined separately as

f(1)=5.

Since the limiting value $2$ is different from the function value $5$ , there is a removable discontinuity, or hole, at $x=1$ .

Step 2: Check $x=2$

Find the one-sided limits.

From the left, use the first rule:

\lim_{x\to 2^-}f(x) = \lim_{x\to 2^-}(x+1) = 3.

From the right, use the last rule:

\lim_{x\to 2^+}f(x) = \lim_{x\to 2^+}\frac{1}{x-3} = -1.

The one-sided limits are different:

3\ne -1.

Thus, the two-sided limit does not exist. Even though $f(2)=4$ is defined, the function has a jump discontinuity at $x=2$ .

Step 3: Check $x=3$

Near $x=3$ , the function is

f(x)=\frac{1}{x-3}.

As $x$ approaches $3$ from the left,

\lim_{x\to 3^-}\frac{1}{x-3}=-\infty,

and from the right,

\lim_{x\to 3^+}\frac{1}{x-3}=+\infty.

Also, $f(3)$ is undefined because the denominator is zero. Therefore, there is an infinite discontinuity, associated with the vertical asymptote

x=3.

Conclusion

The function has:

a removable discontinuity at $x=1$ ,
a jump discontinuity at $x=2$ ,
an infinite discontinuity at $x=3$ .

The main test is to compare the function value with the limit, and when necessary, compare the left-hand and right-hand limits.

Learn by doing: Identify discontinuities

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Limits - Piecewise Function (Graph) to Limit Exists and Is Continuous

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