Precise Definition: Infinite Limits
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.5: “The Precise Definition of a Limit” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: The Challenge Game for Infinite Limits
For $\lim_{x \to 0^+} \dfrac{1}{x} = +\infty$: the output grows without bound as $x \to 0^+$.
The challenge game changes: instead of an opponent asking for output within $\varepsilon$ of $L$, the opponent now names a lower bound $M > 0$ and asks: “Can you guarantee $f(x) > M$?”
Predict: if the opponent says $M = 1000$, how close to 0 must $x$ be for $1/x > 1000$?
$1/x > 1000$ when $x < 1/1000$. So $\delta = 1/1000$ works.
For general $M > 0$: choose $\delta = 1/M$.
Quantity-First Framing
The statement $\lim_{x \to a} f(x) = +\infty$ means: for any threshold $M > 0$ (no matter how large), there exists a distance $\delta > 0$ such that $f(x) > M$ whenever $0 < |x - a| < \delta$.
This captures “grows without bound”: you can make $f(x)$ exceed any specific value, just by getting $x$ close enough to $a$.
Prerequisite Check
Quick Reference
Definition. $\lim_{x \to a} f(x) = +\infty$ means: for every $M > 0$, there exists $\delta > 0$ such that if $0 < |x - a| < \delta$, then $f(x) > M$.
Definition. $\lim_{x \to a} f(x) = -\infty$ means: for every $N < 0$, there exists $\delta > 0$ such that if $0 < |x - a| < \delta$, then $f(x) < N$.
Key Concepts
1. Proving $\lim_{x \to 0} \dfrac{1}{x^2} = +\infty$
Scratch. Given $M > 0$. Need $\dfrac{1}{x^2} > M$ whenever $0 < |x| < \delta$.
$\dfrac{1}{x^2} > M \Leftrightarrow x^2 < \dfrac{1}{M} \Leftrightarrow |x| < \dfrac{1}{\sqrt{M}}$.
Choose $\delta = \dfrac{1}{\sqrt{M}}$.
Formal proof.
Given $M > 0$. Let $\delta = \dfrac{1}{\sqrt{M}}$.
If $0 < |x| < \delta$: $x^2 < \delta^2 = \dfrac{1}{M}$, so $\dfrac{1}{x^2} > M$. $\square$
2. Two Representations
Number-line picture. $M$ is a horizontal line at height $M$ in the output space. The claim is that the graph of $1/x^2$ rises above this line for all $x$ in a punctured disk of radius $\delta = 1/\sqrt{M}$ around 0.
Table. For $\lim_{x \to 0} 1/x^2 = +\infty$:
| $M$ | $\delta = 1/\sqrt{M}$ | Check: $1/\delta^2 = M$ |
|---|---|---|
| 100 | $0.1$ | $1/0.01 = 100$. ✓ |
| 10000 | $0.01$ | $1/0.0001 = 10000$. ✓ |
| $10^6$ | $0.001$ | $10^6$. ✓ |
As $M$ grows (higher threshold), $\delta$ shrinks (must get closer to 0).
3. One-Sided Infinite Limits
For $\lim_{x \to 0^+} 1/x = +\infty$: the definition uses only $0 < x - 0 < \delta$ (right side only).
$\dfrac{1}{x} > M \Leftrightarrow x < \dfrac{1}{M}$. Choose $\delta = \dfrac{1}{M}$.
For $\lim_{x \to 0^-} 1/x = -\infty$: use $-\delta < x < 0$.
$\dfrac{1}{x} < N$ (with $N < 0$) when $x < 0$ and $x > 1/N$ (which is negative). Choose $\delta = -1/N = 1/|N|$.
4. Ask Why: Why Is This Different from the Finite Limit Definition?
In the finite limit definition, the opponent names $\varepsilon > 0$ (an output tolerance), and you name $\delta$ (an input restriction). The output is required to stay NEAR $L$.
In the infinite limit definition, the opponent names $M > 0$ (an output threshold), and you name $\delta$. The output is required to EXCEED $M$ -- to be LARGE, not close to a specific value.
The structure is symmetric: in both cases, the opponent names a challenge (closeness or largeness) and you respond with a $\delta$. The difference is what “success” means for the output.
Named Misconception: asymptote-as-wall
The definition clarifies that the limit $= +\infty$ is a statement about what happens as $x \to a$, not a statement about the value at $a$. The function is not defined at $a = 0$ (for $1/x^2$), and the vertical line $x = 0$ is an asymptote.
An asymptote is not a wall the function “cannot cross.” For $1/x$: the function is positive to the right of 0 and negative to the left -- the function is on BOTH sides of $x = 0$, just discontinuously. The asymptote describes behavior approaching the line, not a barrier preventing the function from crossing it.
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Confusing $M$ and $\varepsilon$ | Writing “Given $\varepsilon > 0$...” for an infinite limit | For $\lim = +\infty$: given $M > 0$; for $\lim = L$: given $\varepsilon > 0$ |
| Wrong inequality direction | $1/x^2 > M \Leftrightarrow x^2 > 1/M$ (wrong) | $1/x^2 > M \Leftrightarrow x^2 < 1/M$ (taking reciprocal flips inequality) |
Leveled Practice
Level 1 -- Finding $\delta$
Problem 1. For $\lim_{x \to 0^+} \dfrac{1}{x} = +\infty$: find $\delta$ in terms of $M$ and verify.
Show answer
Need $1/x > M$: $x < 1/M$. Choose $\delta = 1/M$.
If $0 < x < \delta$: $x < 1/M$, so $1/x > M$. $\square$
Problem 2. Prove $\lim_{x \to 3} \dfrac{1}{(x-3)^2} = +\infty$.
Show answer
Need $\dfrac{1}{(x-3)^2} > M$: $(x-3)^2 < 1/M$, $|x-3| < 1/\sqrt{M}$.
Choose $\delta = 1/\sqrt{M}$. If $0 < |x-3| < \delta$: $\dfrac{1}{(x-3)^2} > M$. $\square$
Level 2 -- Negative Infinite Limits
Problem 3. Prove $\lim_{x \to 0^-} \dfrac{1}{x} = -\infty$.
Show answer
For any $N < 0$: need $1/x < N$ with $x < 0$.
$1/x < N \Leftrightarrow x > 1/N$ (flipping the inequality since $N < 0$ and $x < 0$).
So need $1/N < x < 0$, i.e., $|x| < |1/N| = 1/|N|$. Choose $\delta = 1/|N|$.
If $-\delta < x < 0$: $x$ is negative with $|x| < 1/|N|$, so $1/x < N$. $\square$
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) Explain in plain language why $\lim_{x \to 0} \dfrac{1}{x^2} = +\infty$ but $\lim_{x \to 0} \dfrac{1}{x}$ does not exist.
(b) (Mid) Write a precise definition for $\lim_{x \to +\infty} f(x) = L$ (limit as $x \to +\infty$, finite $L$). Explain how it parallels the $\lim_{x \to a}$ definition.
(c) (Ceiling) Combine definitions: state precisely what it means for $\lim_{x \to +\infty} f(x) = +\infty$. Give an example.
Show answer
(a) $1/x^2 > 0$ for all $x \neq 0$, so it goes to $+\infty$ from both sides. $1/x$ goes to $+\infty$ from the right and $-\infty$ from the left -- the two one-sided limits disagree, so the two-sided limit does not exist.
(b) $\lim_{x \to +\infty} f(x) = L$ means: for every $\varepsilon > 0$, there exists $N > 0$ such that if $x > N$, then $|f(x) - L| < \varepsilon$.
The parallel: instead of restricting $|x - a| < \delta$ (input near $a$), restrict $x > N$ (input very large). The output still must be near $L$.
(c) $\lim_{x \to +\infty} f(x) = +\infty$ means: for every $M > 0$, there exists $N > 0$ such that if $x > N$, then $f(x) > M$.
Example: $\lim_{x \to +\infty} x^2 = +\infty$. Given $M > 0$: choose $N = \sqrt{M}$. If $x > \sqrt{M}$: $x^2 > M$. $\square$
Common Misconceptions
$\lim_{x \to a} f(x) = \infty$ means the function is defined at infinity. The notation names a behavior, not a destination. The formal definition uses a bound $M$: for every $M > 0$, the function eventually exceeds $M$ near $a$. There is no point called “infinity” that the function reaches. Writing $f(a) = \infty$ is undefined; the limit statement describes what happens as $x$ approaches $a$, not at $a$.
a vertical asymptote at $x = a$ means the graph never crosses the line $x = a$. A vertical asymptote marks a value where the function grows without bound; it does not enforce a rule that the graph cannot touch the vertical line at other points or elsewhere. The asymptote describes local behavior near $x = a$, and the function may cross $x = a$ at points not captured by the limit.
Mastery Checklist
Mental Model
The infinite limit definition is the challenge game with a different kind of challenge. For finite limits, the opponent asks “stay within $\varepsilon$ of $L$”; for infinite limits, the opponent asks “exceed $M$.” In both cases, you respond by naming how close $x$ must be to $a$.
The definition captures “without bound”: for any threshold, no matter how large, the function exceeds it. There is no finite value that the function stays near; it escapes to infinity.
Connections
Within MATH161
- Asymptotes: The infinite-limit definition formalizes exactly what a vertical asymptote means: the function exceeds any given threshold near $a$.
- Limits at infinity: The parallel definition for $x \to \infty$ uses a threshold $N$ on the input instead of $\delta$, and a threshold $M$ or $\varepsilon$ on the output.
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