The Formal Definition of One-Sided Limits, calculus lesson

A one-sided limit only looks at half of the neighborhood of a point. The precise version keeps the epsilon-delta structure and changes a single inequality, the one that says which inputs are allowed. That small change is what lets us prove limits at the edge of a domain and at the jumps of piecewise functions.

Key terms

One-sided epsilon-delta definitions

Definition Let $L$ be a real number.

If $f$ is defined on an interval $(a, c)$ with $c > a$ , then $\lim_{x \to a^+} f(x) = L$ means that for every $\varepsilon > 0$ there exists $\delta > 0$ such that

0 < x - a < \delta \implies \lvert f(x) - L \rvert < \varepsilon.

If $f$ is defined on an interval $(c, a)$ with $c < a$ , then $\lim_{x \to a^-} f(x) = L$ means that for every $\varepsilon > 0$ there exists $\delta > 0$ such that

0 < a - x < \delta \implies \lvert f(x) - L \rvert < \varepsilon.

In plain words Same game as before, but only inputs on one side of $a$ are allowed to enter.

Example Here $\lim_{x \to 2^+} (3x - 1) = 5$ . Given $\varepsilon > 0$ , take $\delta = \varepsilon / 3$ . If $0 < x - 2 < \delta$ , then

\lvert (3x - 1) - 5 \rvert = 3(x - 2) < 3\delta = \varepsilon.

Approach from the right

Definition The condition $0 < x - a < \delta$ is equivalent to $a < x < a + \delta$ . The left inequality $0 < x - a$ says $x$ is strictly to the right of $a$ , and the right inequality keeps $x$ within $\delta$ of $a$ . So the right-hand definition tests exactly the inputs in the open interval $(a, a + \delta)$ .

In plain words The allowed inputs form a thin interval just to the right of $a$ , with $a$ itself left out.

Example To prove $\lim_{x \to 0^+} \sqrt{x} = 0$ , let $\varepsilon > 0$ and take $\delta = \varepsilon^2$ . If $0 < x < \varepsilon^2$ , then

\lvert \sqrt{x} - 0 \rvert = \sqrt{x} < \sqrt{\varepsilon^2} = \varepsilon,

because the square root is increasing on $[0, \infty)$ .

Approach from the left

Definition The condition $0 < a - x < \delta$ is equivalent to $a - \delta < x < a$ . It tests exactly the inputs in the open interval $(a - \delta, a)$ .

In plain words The mirror image. The allowed inputs form a thin interval just to the left of $a$ .

Example For the floor function, $\lim_{x \to 1^-} \lfloor x \rfloor = 0$ . Given any $\varepsilon > 0$ , take $\delta = 1$ . If $0 < 1 - x < 1$ , then $0 < x < 1$ , so $\lfloor x \rfloor = 0$ and $\lvert \lfloor x \rfloor - 0 \rvert = 0 < \varepsilon$ . Here one $\delta$ happens to serve every $\varepsilon$ , because the function is constant on that side.

Domain endpoints

Definition If the domain of $f$ is an interval $[a, b]$ with $a < b$ , then at $a$ only the right-hand limit can be formed and at $b$ only the left-hand limit. The two-sided limit is not defined at an endpoint, so the one-sided limit from inside the domain is the statement to prove there. In the looser convention mentioned in the lesson on two-sided limits, the two-sided limit at an endpoint simply equals that one-sided limit.

In plain words At the edge of the domain there is only one direction to come from, so the one-sided definition is the right tool there.

Example For $f(x) = \sqrt{1 - x^2}$ on $[-1, 1]$ we prove $\lim_{x \to 1^-} f(x) = 0$ . Let $\varepsilon > 0$ and take $\delta = \min(1, \varepsilon^2/2)$ . If $0 < 1 - x < \delta$ , then $0 < x < 1$ , so $1 + x < 2$ , and

\sqrt{1 - x^2} = \sqrt{(1 - x)(1 + x)} < \sqrt{\frac{\varepsilon^2}{2} \cdot 2} = \varepsilon.

Proofs and counterexamples

Definition To prove $\lim_{x \to a^+} f(x) = L$ , give a rule that produces a valid $\delta$ for every $\varepsilon > 0$ . To disprove it, use the negation. There exists $\varepsilon > 0$ such that for every $\delta > 0$ some $x$ satisfies

0 < x - a < \delta \quad \text{and} \quad \lvert f(x) - L \rvert \ge \varepsilon.

To show that the right-hand limit does not exist at all, the disproof must work for every real $L$ . The left-hand case is the same with $0 < a - x < \delta$ .

In plain words A proof answers every challenge. A counterexample needs only one challenge that nothing can answer.

Example We show that the right-hand limit of $f(x) = \sin (1/x)$ at $0$ does not exist. Let $L$ be any real number and take $\varepsilon = 1/2$ . For any $\delta > 0$ , choose a positive whole number $k$ so large that the two inputs

x_1 = \frac{1}{\pi/2 + 2k\pi} \quad \text{and} \quad x_2 = \frac{1}{3\pi/2 + 2k\pi}

both lie in $(0, \delta)$ . Then $f(x_1) = 1$ and $f(x_2) = -1$ . If both were within $1/2$ of $L$ , the triangle inequality would give

2 = \lvert 1 - (-1) \rvert \le \lvert 1 - L \rvert + \lvert L - (-1) \rvert < \frac{1}{2} + \frac{1}{2} = 1,

which is false. So for each $\delta$ at least one of $x_1, x_2$ satisfies $\lvert f(x) - L \rvert \ge 1/2$ .

Check yourself

Prove that $\lim_{x \to 3^+} \sqrt{x - 3} = 0$ .

Show the answer

Let $\varepsilon > 0$ and take $\delta = \varepsilon^2$ . If $0 < x - 3 < \varepsilon^2$ , then $\sqrt{x - 3} < \varepsilon$ , because the square root is increasing.

Use the definition to show that

\lim_{x \to 0^-} \frac{\lvert x \rvert}{x} = -1.

Show the answer

For any $\varepsilon > 0$ take $\delta = 1$ (any positive number works). If $0 < 0 - x < 1$ , then $x < 0$ , so $\lvert x \rvert / x = -1$ and the distance to $-1$ is $0 < \varepsilon$ .

Suppose $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = L$ . Prove that $\lim_{x \to a} f(x) = L$ .

Show the answer

Let $\varepsilon > 0$ . The left-hand limit gives $\delta_1 > 0$ that works for inputs in $(a - \delta_1, a)$ , and the right-hand limit gives $\delta_2 > 0$ that works for inputs in $(a, a + \delta_2)$ . Take $\delta = \min(\delta_1, \delta_2)$ . Any $x$ with $0 < \lvert x - a \rvert < \delta$ lies in one of those two intervals, so $\lvert f(x) - L \rvert < \varepsilon$ .

The Formal Definition of One-Sided Limits

Theory, part 11

Practice, part 12

Key terms

One-sided epsilon-delta definitions

Approach from the right

Approach from the left

Domain endpoints

Proofs and counterexamples

Check yourself

How this lesson connects

Builds on