Left and Right Limit Method

Almost every continuity question you will ever be asked is really a question about a piecewise function: a formula that changes at one particular point. The temptation is to plug the point into the formula and declare victory, but that only works if you happened to guess the right branch. The left-right limit method removes the guesswork. It asks three separate questions — what happens as you approach from the left, what happens as you approach from the right, and what is the function actually defined to be at the point — and then compares the three answers.

That comparison is more informative than it looks. When the three agree you have continuity; when the two limits agree but differ from the function value you have a removable discontinuity; when the two limits both exist but differ you have a jump; and when a one-sided limit fails to exist at all you have a discontinuity of the second kind. This page sets out the method step by step, works through three examples, and then addresses the question that comes up as soon as you start using it seriously: how do you know a one-sided limit exists at all?

The Principle of the Test

A function $f(x)$ is continuous at $x_0$ if and only if

$$\lim_{x \to x_0^-} f(x) = \lim_{x \to x_0^+} f(x) = f(x_0)$$

The three quantities in that chain are genuinely independent, and the test is the conjunction of three conditions:

1. The left-hand limit exists and is finite.
2. The right-hand limit exists and is finite.
3. Both limits and the function value are the same number.

The Four Steps

1. Compute the left limit: $\lim_{x \to x_0^-} f(x)$, by taking the branch of the formula that applies for $x$ slightly less than $x_0$.
2. Compute the right limit: $\lim_{x \to x_0^+} f(x)$, using the branch that applies for $x$ slightly greater than $x_0$.
3. Compute the function value: $f(x_0)$, using whichever branch actually contains $x_0$. If no branch contains $x_0$, the function value is simply undefined, and that is already a failure.
4. Compare all three: if they are equal, $f$ is continuous at $x_0$; if not, it is not.

Step 3 is where students most often slip. The branch that governs the limit from the left is frequently not the branch that defines the value at the point, and the method exists precisely to force you to check both.

A Comparison Table of the Outcomes

The table is a decision procedure. The first row is the only one that gives continuity; the second row is the case that a limits-only test would miss; the last three rows are all second-kind discontinuities, distinguished by which side fails.

Worked Examples

Example 1. Determine the continuity of $f(x) = \begin{cases} x^2, & x \leq 1 \\ 2x - 1, & x > 1 \end{cases}$ at $x = 1$.

Solution:

1. Left limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1$.
2. Right limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 1$.
3. Function value: $f(1) = 1^2 = 1$, using the branch that contains $x = 1$.
4. All three are equal.
5. Conclusion: the function is continuous at $x = 1$.

Example 2. Determine the continuity of $f(x) = \begin{cases} x, & x < 0 \\ x + 1, & x \geq 0 \end{cases}$ at $x = 0$.

Solution:

1. Left limit: $\lim_{x \to 0^-} f(x) = 0$.
2. Right limit: $\lim_{x \to 0^+} f(x) = 1$.
3. Function value: $f(0) = 1$, since $0$ belongs to the second branch.
4. The left limit and the right limit are not equal.
5. Conclusion: the function is discontinuous at $x = 0$, and the discontinuity is a jump — the graph has an open circle at height $0$ and a filled circle at height $1$.

Example 3. Determine the continuity of $f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \neq 1 \\ 0, & x = 1 \end{cases}$ at $x = 1$.

Solution:

1. Left limit: for $x$ near $1$ with $x \neq 1$, the factorisation $x^2 - 1 = (x-1)(x+1)$ gives $\frac{x^2-1}{x-1} = x + 1$, so $\lim_{x \to 1^-} f(x) = 2$.
2. Right limit: by the same computation, $\lim_{x \to 1^+} f(x) = 2$.
3. Function value: $f(1) = 0$, by definition.
4. The two limits agree with each other but not with the function value.
5. Conclusion: the function is discontinuous at $x = 1$, and the discontinuity is removable — redefining $f(1) = 2$ would make the function continuous.

Do One-Sided Limits Always Exist?

The method above assumes you can compute $\lim_{x \to x_0^-} f(x)$ and $\lim_{x \to x_0^+} f(x)$. Sometimes you cannot, and then the honest answer is that the limit does not exist — which is itself a conclusion, and one that the comparison table above is designed to absorb. Three standard results tell you when to expect success.

The two-sided criterion. The limit $\lim_{x \to x_0} f(x)$ exists if and only if both one-sided limits exist and are equal. This is the reason the method is named after the two sides: working one-sided is always easier, because on one side of $x_0$ a piecewise function usually reduces to a single smooth formula, and the one-sided limit is then just the ordinary limit of that formula.

The squeeze theorem (sandwich theorem). Suppose that near $x_0$ you can find two functions $g$ and $h$ with $g(x) \leq f(x) \leq h(x)$, and that $\lim_{x \to x_0^-} g(x) = \lim_{x \to x_0^-} h(x) = L$. Then $\lim_{x \to x_0^-} f(x) = L$ as well. The same statement holds on the right. This is the standard tool for one-sided limits of expressions such as $\frac{\sin x}{x}$ or $\frac{1 - \cos x}{x^2}$, where direct substitution produces the indeterminate form $\frac{0}{0}$ and no algebraic simplification is available.

The monotone bounded theorem. A monotone increasing sequence that is bounded above converges, and a monotone decreasing sequence that is bounded below converges. The continuous-function analogue is this: a function that is monotone on an interval has one-sided limits (possibly infinite) at every interior point of that interval. In practice this means that for a monotone branch of a piecewise function you can be confident the one-sided limit exists before you start computing it, and you can use the values from the correct side to identify it.

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Exercises

Exercise 1

Use the left and right limit method to determine the continuity of $f(x) = \begin{cases} x^2, & x < 0 \\ x + 1, & x \geq 0 \end{cases}$ at $x = 0$.

Exercise 2

Determine the continuity of $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$, and classify the type of the discontinuity.

Exercise 3

Let $f$ be continuous on $[0, 1]$ with $f(0) = 0$ and $f(1) = 1$. Show that there is a point $c \in (0, 1)$ with $f(c) = \frac{1}{2}$.

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Summary

The left-right limit method is the workhorse test for continuity at a point where the formula changes. It reduces every question to three numbers — the limit from the left, the limit from the right, and the value at the point — and then reads the conclusion off their pattern: all three equal means continuous; two equal and one different means a removable discontinuity; two limits that differ means a jump; a missing one-sided limit means a discontinuity of the second kind. What makes the method practical is that one-sided limits are usually easier to compute than two-sided ones, since on each side a piecewise function reduces to a single smooth formula, and the squeeze theorem and the monotone bounded theorem tell you in advance whether the limit you are chasing actually exists.

Symbols Used in This Article