The Timpson Lyceum · Calculus 1 (2024)

Definition of Limits & One-Sided Limits - Lecture 5 (Calculus 1)

Overview

This lecture replaces the intuitive "what the function gets close to" idea of a limit with the precise epsilon–delta ( $\varepsilon$ – $\delta$ ) definition, the foundational definition of calculus. Because continuity, derivatives, and integrals are all built on limits, mastering this definition is essential. After motivating the definition with an interval "game," we work through several $\varepsilon$ – $\delta$ proofs (including the limit of a sum), then introduce one-sided (left-hand and right-hand) limits and the theorem connecting them to two-sided limits. The lecture closes with the squeeze (sandwich) theorem and its use to evaluate trigonometric limits like $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}$ .

Key Concepts

The interval "game" (intuition). Given $f$ , a point $c$ , and a target output $L$ , the goal is: for any chosen tolerance around $L$ , find an interval around $c$ whose image lands entirely inside that tolerance. You only need to find an interval that works — not the largest one. The output tolerance is $\varepsilon$ ; the input radius is $\delta$ .

Precise definition of a limit. Let $f(x)$ be defined on an open interval around $c$ , except possibly at $c$ itself. We say $\lim_{x \to c} f(x) = L$ provided that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all $x$ , $0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon.$

Why $|f(x) - L|$ (a difference). The absolute value of a difference $|a - b|$ is the distance between two numbers; we want the output to stay within $\varepsilon$ of $L$ .
Why $0 < |x - c|$ . The strict lower bound excludes $x = c$ . A limit never cares about the value (or even existence) of $f$ at $c$ — only about behavior arbitrarily close to $c$ . This is why a function can have a hole at $c$ and still have a limit there.
$\implies$ is not $=$ . " $P \implies Q$ " means $P$ being true forces $Q$ ; it does not mean $P$ and $Q$ are equal.

How an $\varepsilon$ – $\delta$ proof works. You are given an arbitrary $\varepsilon > 0$ and must produce a $\delta$ (expressed in terms of $\varepsilon$ ) that makes the implication hold. There is no single "right" $\delta$ — infinitely many work, since any smaller $\delta$ also works. The standard method:

Scratch work (find $\delta$ ): start from $|f(x) - L| < \varepsilon$ , substitute $f(x)$ , and algebraically solve until you isolate $|x - c|$ . The bound on $|x - c|$ that emerges is your $\delta$ .
The proof (verify $\delta$ ): declare your $\delta$ , assume $|x - c| < \delta$ , and show the chain of implications runs forward to $|f(x) - L| < \varepsilon$ . The logic flows so the input bound implies the output bound.

The $\min$ trick for nonlinear $f$ . When simplifying produces an extra variable factor (e.g. $|x - 2|\cdot|x + 2|$ ), you cannot bound $|x+2|$ until you first restrict $x$ . Set $\delta = \min\{1,\ \text{(expression in } \varepsilon)\}$ . The " $1$ " pre-bounds $x$ to a finite interval, giving a numeric upper bound on the troublesome factor; the $\varepsilon$ -expression then forces $|f(x)-L| < \varepsilon$ .

Triangle inequality. $|a + b| \le |a| + |b|$ . This is the key tool for proving the limit of a sum: group $f(x)+g(x)-(L+M)$ as $(f(x)-L)+(g(x)-M)$ and bound each piece by $\varepsilon/2$ .

One-sided limits. A one-sided limit approaches $c$ from only one direction.

Right-hand limit: $\displaystyle\lim_{x \to c^+} f(x) = L$ if for every $\varepsilon > 0$ there is a $\delta > 0$ such that $c < x < c + \delta \implies |f(x) - L| < \varepsilon$ (uses only points to the right of $c$ ).
Left-hand limit: $\displaystyle\lim_{x \to c^-} f(x) = L$ uses $c - \delta < x < c \implies |f(x) - L| < \varepsilon$ (points to the left).
The definitions are identical to the two-sided one except the input condition restricts the side; $x = c$ is still excluded.

Existence theorem. The two-sided limit exists and equals $L$ if and only if both one-sided limits exist and agree: $\lim_{x \to c} f(x) = L \iff \lim_{x \to c^+} f(x) = L = \lim_{x \to c^-} f(x).$ If the one-sided limits differ (e.g. a jump), no two-sided limit exists. At a domain endpoint (e.g. $\sqrt{4 - x^2}$ at $x = 2$ , or $\sqrt{x}$ at $x = 0$ ), one side is undefined, so only a one-sided limit can exist.

Limits need not exist. A function defined "nicely" can still fail to have a limit: $\sin(1/x)$ oscillates faster and faster as $x \to 0$ , never settling on one value, so $\lim_{x\to 0}\sin(1/x)$ does not exist.

Squeeze (Sandwich) Theorem. If $g(x) \le f(x) \le h(x)$ near $c$ and $\lim_{x\to c} g(x) = \lim_{x\to c} h(x) = L$ , then $\lim_{x\to c} f(x) = L$ . Using the unit-circle area comparison $\text{Area}(\triangle OAB) < \text{Area(sector } OAB) < \text{Area}(\triangle OCB)$ (the sector area being $\tfrac12\theta r^2$ for radius $r=1$ ) gives, after simplifying, the central trig limit: $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1.$

Companion trig limits. Built from the result above and the limit laws (the limit of a product/constant multiple distributes), e.g. $\lim_{\theta\to0}\frac{1-\cos\theta}{\theta} = 0$ , $\lim_{x\to0}\frac{\sin 2x}{5x} = \frac{2}{5}$ , and $\lim_{t\to0}\frac{\tan t\,\sec 2t}{3t} = \frac13$ .

Worked Example

Prove that $\displaystyle\lim_{x \to 1}(5x - 3) = 2$ .

Here $f(x) = 5x - 3$ , $c = 1$ , $L = 2$ . Given any $\varepsilon > 0$ , work backward from $|f(x) - L| < \varepsilon$ to find $\delta$ , then state the forward proof.

Rule (what must be shown): $0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon$ .

\begin{aligned} |f(x) - 2| &= |(5x - 3) - 2| = |5x - 5| = 5\,|x - 1| \\ |f(x) - 2| < \varepsilon &\iff 5\,|x - 1| < \varepsilon \iff |x - 1| < \tfrac{\varepsilon}{5} \quad\Rightarrow\quad \text{let } \delta = \tfrac{\varepsilon}{5} \\ \text{Then } |x - 1| < \delta &\implies |f(x) - 2| = 5\,|x - 1| < 5\delta = 5\cdot\tfrac{\varepsilon}{5} = \varepsilon. \end{aligned}

Since for every $\varepsilon > 0$ the choice $\delta = \varepsilon/5$ forces $|f(x) - 2| < \varepsilon$ , the limit is $2$ . $\blacksquare$

Summary

The precise definition: $\lim_{x\to c} f(x)=L$ means for every $\varepsilon>0$ there is a $\delta>0$ with $0<|x-c|<\delta \implies |f(x)-L|<\varepsilon$ — the input bound $\delta$ guarantees the output bound $\varepsilon$ .
In a proof you are handed $\varepsilon$ and must construct a $\delta$ (in terms of $\varepsilon$ ); any sufficiently small $\delta$ works, so there is no unique answer.
For nonlinear functions, use $\delta = \min\{1, \text{expr}(\varepsilon)\}$ to first bound the extra variable factor, and use the triangle inequality $|a+b|\le|a|+|b|$ to prove the limit of a sum (split the error into $\varepsilon/2$ pieces).
One-sided limits restrict the approach to $x > c$ (right) or $x < c$ (left); the two-sided limit exists iff both one-sided limits exist and are equal.
A limit need not exist (e.g. $\sin(1/x)$ at $0$ ), and at domain endpoints only a one-sided limit is meaningful.
The Squeeze Theorem traps $f$ between two functions with a common limit; it yields $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ , the foundation for the other trigonometric limits.