Epsilon-Delta Variations: One-Sided and Infinite Limits

MATH161

Reference: Stewart 2.4 • Chapter: 1 • Section: 4

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Epsilon-Delta Variations: One-Sided and Infinite Limits

Beyond the Standard Definition

The standard epsilon-delta definition handles $\lim_{x \to a} f(x) = L$ where $x$ approaches $a$ from both sides and $L$ is a finite number. But what about:

One-sided limits: $x$ approaches only from the left ($x \to a^-$) or right ($x \to a^+$)
Infinite limits: The limit "equals" $\infty$ or $-\infty$
Limits at infinity: $x \to \infty$ or $x \to -\infty$

Each variation follows the same logical structure—just change what we're controlling and what we're guaranteeing.

Prerequisite Map

Quick Reference

Property	Value
Course	MATH161
Chapter.Section	1.7
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

Comparison of All Definitions

Limit Type	Definition
Two-sided: $\lim_{x \to a} f(x) = L$	$\forall \varepsilon > 0,\ \exists \delta > 0:\ 0 < \\|x - a\\| < \delta \Rightarrow \\|f(x) - L\\| < \varepsilon$
Left-hand: $\lim_{x \to a^-} f(x) = L$	$\forall \varepsilon > 0,\ \exists \delta > 0:\ a - \delta < x < a \Rightarrow \\|f(x) - L\\| < \varepsilon$
Right-hand: $\lim_{x \to a^+} f(x) = L$	$\forall \varepsilon > 0,\ \exists \delta > 0:\ a < x < a + \delta \Rightarrow \\|f(x) - L\\| < \varepsilon$
Infinite: $\lim_{x \to a} f(x) = \infty$	$\forall M > 0,\ \exists \delta > 0:\ 0 < \\|x - a\\| < \delta \Rightarrow f(x) > M$
At infinity: $\lim_{x \to \infty} f(x) = L$	$\forall \varepsilon > 0,\ \exists N > 0:\ x > N \Rightarrow \\|f(x) - L\\| < \varepsilon$

One-Sided Limits

Left-hand limit: $\lim_{x \to a^-} f(x) = L$

$$\forall \varepsilon > 0,\ \exists \delta > 0 \text{ such that } a - \delta < x < a \Rightarrow \vert f(x) - L\vert < \varepsilon$$

        f(x)
          │
    L+ε ──┼────────────────
          │    ●●●●●●
      L ──┼───●───────────○  (open at x=a)
          │  ●
    L-ε ──┼─●─────────────
          │
          └───────────┼───→ x
                a-δ   a
              ◄───────►
              only x < a

Right-hand limit: $\lim_{x \to a^+} f(x) = L$

$$\forall \varepsilon > 0,\ \exists \delta > 0 \text{ such that } a < x < a + \delta \Rightarrow \vert f(x) - L\vert < \varepsilon$$

        f(x)
          │
    L+ε ──┼────────────────
          │         ●●●●●●
      L ──┼───○────●───────  (open at x=a)
          │         ●
    L-ε ──┼──────────●─────
          │
          └─────┼───────┼──→ x
                a     a+δ
                ◄───────►
                only x > a

Key difference: Replace "$0 < \vert x - a\vert < \delta$" with:

Left: "$a - \delta < x < a$" (x is less than a)
Right: "$a < x < a + \delta$" (x is greater than a)

Infinite Limits

Definition: $\lim_{x \to a} f(x) = \infty$

$$\forall M > 0,\ \exists \delta > 0 \text{ such that } 0 < \vert x - a\vert < \delta \Rightarrow f(x) > M$$

Key changes:

Replace $\varepsilon$ (small tolerance) with $M$ (large threshold)
Replace "$\vert f(x) - L\vert < \varepsilon$" with "$f(x) > M$"
The "game" is now: no matter how large an $M$ you demand, I can make $f(x)$ exceed it by getting $x$ close enough to $a$

    f(x)
      │             │
      │             │
    M ├─────────────┼─────
      │           ╱ │ ╲
      │         ╱   │   ╲
      │       ╱     │     ╲
      │     ╱       │       ╲
      └─────────────┼─────────→ x
                    a
                 ◄──┴──►
                   δ

For $\lim_{x \to a} f(x) = -\infty$: Change "$f(x) > M$" to "$f(x) < -M$".

Limits at Infinity

Definition: $\lim_{x \to \infty} f(x) = L$

$$\forall \varepsilon > 0,\ \exists N > 0 \text{ such that } x > N \Rightarrow \vert f(x) - L\vert < \varepsilon$$

Key changes:

Replace $\delta$ with $N$ (a large threshold for $x$)
Replace "$0 < \vert x - a\vert < \delta$" with "$x > N$"
We're saying: for any $\varepsilon$, making $x$ large enough keeps $f(x)$ within $\varepsilon$ of $L$

    f(x)
      │
  L+ε ├───────────────●●●●●●●●●
      │            ●●●
    L ├──────────●●────────────
      │        ●●
  L-ε ├──────●●────────────────
      │    ●●
      │  ●●
      └─────┼─────────────────→ x
            N
            ◄─────────────────►
                  x > N

Worked Example: One-Sided Limit

Prove: $\lim_{x \to 0^+} \sqrt{x} = 0$

Proof: Let $\varepsilon > 0$ be given. Choose $\delta = \varepsilon^2$.

Suppose $0 < x < \delta = \varepsilon^2$.

Then $\sqrt{x} < \sqrt{\varepsilon^2} = \varepsilon$ (since $\sqrt{}$ is increasing).

So $\vert \sqrt{x} - 0\vert = \sqrt{x} < \varepsilon$.

Therefore $\lim_{x \to 0^+} \sqrt{x} = 0$. $\square$

Worked Example: Infinite Limit

Prove: $\lim_{x \to 0} \frac{1}{x^2} = \infty$

Proof: Let $M > 0$ be given. Choose $\delta = \frac{1}{\sqrt{M}}$.

Suppose $0 < \vert x\vert < \delta = \frac{1}{\sqrt{M}}$.

Then $\vert x\vert ^2 < \frac{1}{M}$, so $\frac{1}{x^2} = \frac{1}{\vert x\vert ^2} > M$.

Therefore $\lim_{x \to 0} \frac{1}{x^2} = \infty$. $\square$

Practice Problems

Level 1 Identify the Definition

Match each limit statement with the correct formal definition:

$\lim_{x \to 3^+} f(x) = 7$
$\lim_{x \to 3} f(x) = \infty$
$\lim_{x \to \infty} f(x) = 7$

Definitions:

$\forall \varepsilon > 0,\ \exists N > 0:\ x > N \Rightarrow \vert f(x) - 7\vert < \varepsilon$
$\forall \varepsilon > 0,\ \exists \delta > 0:\ 3 < x < 3 + \delta \Rightarrow \vert f(x) - 7\vert < \varepsilon$
$\forall M > 0,\ \exists \delta > 0:\ 0 < \vert x - 3\vert < \delta \Rightarrow f(x) > M$

Thought Process

(a) is a right-hand limit approaching a finite value
(b) is an infinite limit (function goes to infinity near $x = 3$)
(c) is a limit at infinity (function approaches 7 as $x$ grows)

Show Answer

Matches **Definition 2** (right-hand limit: $3 < x < 3 + \delta$)
Matches **Definition 3** (infinite limit: $f(x) > M$)
Matches **Definition 1** (limit at infinity: $x > N$)

Level 2 One-Sided Limit Proof

Prove that $\lim_{x \to 4^-} (2x + 1) = 9$ using the epsilon-delta definition for left-hand limits.

Thought Process

This is just like a regular linear limit proof, except we only consider $x < 4$.

Need: $\vert (2x + 1) - 9\vert = \vert 2x - 8\vert = 2\vert x - 4\vert < \varepsilon$

So $\vert x - 4\vert < \varepsilon/2$, meaning $-\varepsilon/2 < x - 4 < \varepsilon/2$.

For left-hand limit, we need $4 - \delta < x < 4$, which is the left part of this interval.

Choose $\delta = \varepsilon/2$.

Show Answer

Proof: Let $\varepsilon > 0$ be given. Choose $\delta = \frac{\varepsilon}{2}$.

Suppose $4 - \delta < x < 4$ (i.e., $x$ is to the left of 4 but within $\delta$).

This means $\vert x - 4\vert < \delta = \frac{\varepsilon}{2}$.

Then: $$\vert (2x + 1) - 9\vert = \vert 2x - 8\vert = 2\vert x - 4\vert < 2 \cdot \frac{\varepsilon}{2} = \varepsilon$$

Therefore $\lim_{x \to 4^-} (2x + 1) = 9$. $\square$

Level 3 Infinite Limit Proof

Prove that $\lim_{x \to 2} \frac{1}{(x-2)^2} = \infty$ using the formal definition.

Thought Process

We need: for every $M > 0$, find $\delta > 0$ such that $0 < \vert x - 2\vert < \delta$ implies $\frac{1}{(x-2)^2} > M$.

Working backwards: $\frac{1}{(x-2)^2} > M$ means $(x-2)^2 < \frac{1}{M}$, so $\vert x - 2\vert < \frac{1}{\sqrt{M}}$.

Choose $\delta = \frac{1}{\sqrt{M}}$.

Show Answer

Proof: Let $M > 0$ be given. Choose $\delta = \frac{1}{\sqrt{M}}$.

Suppose $0 < \vert x - 2\vert < \delta$.

Then: $$\vert x - 2\vert ^2 < \delta^2 = \frac{1}{M}$$

So: $$\frac{1}{(x-2)^2} = \frac{1}{\vert x-2\vert ^2} > M$$

Therefore $\lim_{x \to 2} \frac{1}{(x-2)^2} = \infty$. $\square$

Level 4 Limit at Infinity

Prove that $\lim_{x \to \infty} \frac{3x + 1}{x} = 3$ using the formal definition.

Thought Process

First simplify: $\frac{3x + 1}{x} = 3 + \frac{1}{x}$

So $\left\vert \frac{3x + 1}{x} - 3\right\vert = \left\vert \frac{1}{x}\right\vert = \frac{1}{x}$ (for $x > 0$).

Need $\frac{1}{x} < \varepsilon$, i.e., $x > \frac{1}{\varepsilon}$.

Choose $N = \frac{1}{\varepsilon}$.

Show Answer

Proof: Let $\varepsilon > 0$ be given. Choose $N = \frac{1}{\varepsilon}$.

Suppose $x > N$.

Then: $$\left\vert \frac{3x + 1}{x} - 3\right\vert = \left\vert 3 + \frac{1}{x} - 3\right\vert = \frac{1}{x} < \frac{1}{N} = \varepsilon$$

Therefore $\lim_{x \to \infty} \frac{3x + 1}{x} = 3$. $\square$

Level 5 Connecting One-Sided and Two-Sided

Prove: $\lim_{x \to a} f(x) = L$ if and only if both $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = L$.

Prove the "if" direction: assuming both one-sided limits equal $L$, show the two-sided limit is $L$.
Prove the "only if" direction: assuming the two-sided limit is $L$, show both one-sided limits are $L$.

Thought Process

For (a): If both one-sided limits give $\delta_1$ and $\delta_2$ respectively, take $\delta = \min\{\delta_1, \delta_2\}$ to cover both sides.

For (b): If the two-sided $\delta$ works for all $x$ with $0 < \vert x - a\vert < \delta$, it certainly works for the subset where $x < a$ or $x > a$.

Show Answer

Part (a): "If" direction

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

Let $\varepsilon > 0$ be given.

By the left-hand limit: $\exists \delta_1 > 0$ such that $a - \delta_1 < x < a \Rightarrow \vert f(x) - L\vert < \varepsilon$.

By the right-hand limit: $\exists \delta_2 > 0$ such that $a < x < a + \delta_2 \Rightarrow \vert f(x) - L\vert < \varepsilon$.

Let $\delta = \min\{\delta_1, \delta_2\}$.

If $0 < \vert x - a\vert < \delta$, then either:

$a - \delta < x < a$ (left side): $\vert f(x) - L\vert < \varepsilon$ by the first condition
$a < x < a + \delta$ (right side): $\vert f(x) - L\vert < \varepsilon$ by the second condition

Therefore $\lim_{x \to a} f(x) = L$. $\square$

Part (b): "Only if" direction

Assume $\lim_{x \to a} f(x) = L$.

Let $\varepsilon > 0$ be given.

Then $\exists \delta > 0$ such that $0 < \vert x - a\vert < \delta \Rightarrow \vert f(x) - L\vert < \varepsilon$.

For the left-hand limit: If $a - \delta < x < a$, then $0 < \vert x - a\vert < \delta$, so $\vert f(x) - L\vert < \varepsilon$.

This proves $\lim_{x \to a^-} f(x) = L$.

For the right-hand limit: If $a < x < a + \delta$, then $0 < \vert x - a\vert < \delta$, so $\vert f(x) - L\vert < \varepsilon$.

This proves $\lim_{x \to a^+} f(x) = L$. $\square$

Conceptual Check (CCI-Style)

Question: For which of the following can we write a formal epsilon-delta style definition?

$\lim_{x \to 5} f(x) = 3$
$\lim_{x \to 5} f(x) = \infty$
$\lim_{x \to \infty} f(x) = 3$
$\lim_{x \to \infty} f(x) = \infty$
All of the above

Answer

(E) All of the above. Each has a precise definition:

(A): Standard $\varepsilon$-$\delta$
(B): Replace $\varepsilon$ with $M$ and "$\vert f(x) - L\vert < \varepsilon$" with "$f(x) > M$"
(C): Replace $\delta$ with $N$ and "$\vert x - a\vert < \delta$" with "$x > N$"
(D): Both replacements: for all $M > 0$, there exists $N > 0$ such that $x > N \Rightarrow f(x) > M$

Summary Table

What Changes	Two-Sided	One-Sided	Infinite	At Infinity
Input condition	$0 < \\|x - a\\| < \delta$	$a < x < a + \delta$ or $a - \delta < x < a$	$0 < \\|x - a\\| < \delta$	$x > N$
Output condition	$\\|f(x) - L\\| < \varepsilon$	$\\|f(x) - L\\| < \varepsilon$	$f(x) > M$	$\\|f(x) - L\\| < \varepsilon$
For every...	$\varepsilon > 0$	$\varepsilon > 0$	$M > 0$	$\varepsilon > 0$
There exists...	$\delta > 0$	$\delta > 0$	$\delta > 0$	$N > 0$

Mastery Checklist

[ ] I can state the definition for left-hand limits ($x \to a^-$)
[ ] I can state the definition for right-hand limits ($x \to a^+$)
[ ] I can state the definition for infinite limits ($f(x) \to \infty$)
[ ] I can state the definition for limits at infinity ($x \to \infty$)
[ ] I can prove one-sided limits using the modified definition
[ ] I can prove infinite limits by finding $\delta$ in terms of $M$
[ ] I can prove that two-sided limits equal both one-sided limits

Mental Model

The Universal Structure:

Every epsilon-delta variation has the same logical skeleton:

"For every challenge (small $\varepsilon$ or large $M$), there exists a response ($\delta$ or $N$) such that satisfying the input condition guarantees the output condition."

Variation	Challenge	Response	Input	Output
Standard	Small $\varepsilon$	Small $\delta$	$x$ near $a$	$f(x)$ near $L$
Infinite	Large $M$	Small $\delta$	$x$ near $a$	$f(x)$ exceeds $M$
At infinity	Small $\varepsilon$	Large $N$	$x$ exceeds $N$	$f(x)$ near $L$

Connections

Looking back:

Epsilon-Delta Definition is the foundation that these variations modify.
One-Sided Limits gave the intuition; now we have precision.

Looking ahead:

These definitions prove that vertical asymptotes occur where $\lim_{x \to a} f(x) = \pm\infty$.
Horizontal asymptotes are characterized by $\lim_{x \to \pm\infty} f(x) = L$.

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Last updated: 2026-01-22