How Can a Limit Fail to Exist?
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.2: “The Limit of a Function” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
Most limits students compute in Calculus I either exist or fail in one of three well-understood ways. Knowing the three failure modes is as important as knowing the success case, because it tells you exactly what went wrong and why no finite limit can be assigned.
The three failures are: the left and right one-sided limits exist but disagree (jump), the function blows up in magnitude near the point (vertical asymptote), and the function oscillates without settling (oscillation). Each leaves a distinctive signature in a table or graph, and each has a specific geometric picture.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
A limit $\lim_{x \to a} f(x)$ fails to exist for exactly one of the following reasons:
| Failure mode | Signature | Example |
|---|---|---|
| Jump | $\lim_{x\to a^-} f(x) \neq \lim_{x\to a^+} f(x)$; both finite | $|x|/x$ at $x=0$ |
| Vertical asymptote | At least one one-sided limit is $\pm\infty$ | $1/x$ at $x=0$ |
| Oscillation | Function bounces between values, settles on nothing | $\sin(1/x)$ at $x=0$ |
Key Concepts
1. Failure Mode 1: Jump (Different Finite One-Sided Limits)
A jump occurs when $f(x)$ settles to one finite value approaching from the left and a different finite value approaching from the right. The function “jumps” at $x = a$ in the graph.
The two-sided limit does not exist, but both one-sided limits are well-defined finite numbers. You can still say something useful: the left limit and the right limit both exist.
Example 1. Consider $f(x) = \dfrac{|x|}{x}$ at $x = 0$.
For $x > 0$: $|x|/x = 1$. Right limit $= 1$.
For $x < 0$: $|x|/x = -1$. Left limit $= -1$.
Since $1 \neq -1$, $\lim_{x \to 0} f(x)$ does not exist. This is a jump.
Example 2. The floor function $f(x) = \lfloor x \rfloor$ at any integer $n$:
Left limit $= n - 1$ (approaching from values just below $n$).
Right limit $= n$ (for $x$ just above $n$, $\lfloor x \rfloor = n$).
Since $n - 1 \neq n$, the two-sided limit fails to exist at every integer. The graph shows a staircase with jumps.
2. Failure Mode 2: Vertical Asymptote (Unbounded)
A vertical asymptote occurs when $|f(x)|$ grows without bound as $x \to a$. In this case, the limit does not exist because no finite number can be the limit. We use $\pm\infty$ to describe the behavior.
The formal statement is: $\lim_{x \to a} f(x) = \infty$ means for every positive number $M$, there exists $\delta > 0$ such that if $0 < |x - a| < \delta$ then $f(x) > M$. The limit is not $\infty$ -- $\infty$ is not a number. Rather, the limit fails to exist, and the failure is of the unbounded type.
Example 3. $\displaystyle\lim_{x \to 0} \frac{1}{x^2}$.
For $x$ near 0: $x^2$ is small and positive, so $1/x^2$ is large and positive. As $x \to 0$ from either side, $1/x^2 \to +\infty$. The limit does not exist (unbounded failure).
Example 4. $\displaystyle\lim_{x \to 2} \frac{1}{(x-2)^2}$.
Near $x = 2$: $(x-2)^2 \to 0^+$, so $1/(x-2)^2 \to +\infty$. Both one-sided limits are $+\infty$. Limit does not exist.
3. Failure Mode 3: Oscillation (No Settlement)
Oscillation is the subtlest failure. The function $f(x)$ fluctuates between multiple values as $x \to a$ without ever settling to any single value, finite or infinite. No table of nearby values will converge; different sequences of $x$-values approaching $a$ give different limiting values.
Example 5. $\displaystyle\lim_{x \to 0} \sin\!\left(\frac{1}{x}\right)$.
As $x \to 0$, $1/x \to \pm\infty$, and $\sin(1/x)$ oscillates between $-1$ and $1$ infinitely often. For example:
- At $x = \frac{2}{\pi}$: $\sin(1/x) = \sin(\pi/2) = 1$.
- At $x = \frac{2}{3\pi}$: $\sin(1/x) = \sin(3\pi/2) = -1$.
- At $x = \frac{1}{n\pi}$ for integer $n$: $\sin(1/x) = 0$.
Every value in $[-1, 1]$ is approached by some sequence of $x$-values tending to 0. No single number is the limit. The limit does not exist (oscillation failure).
4. Distinguishing the Three Failures
When asked “Does $\lim_{x \to a} f(x)$ exist?”, proceed systematically:
- Compute $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$.
- If both are finite and equal: limit exists.
- If both are finite but unequal: jump failure.
- If at least one is $\pm\infty$: unbounded failure.
- If the function oscillates (values approach $a$ but $f(x)$ never settles): oscillation failure.
For standard functions in Calculus I, you will most often encounter jumps (from piecewise functions and $|f(x)|$) and vertical asymptotes (from rational functions, $\ln$, and $\tan$). Oscillation appears in special examples like $\sin(1/x)$ that illuminate why the definition of limit requires every sequence to converge, not just nice sequences.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Thinking “$\infty$” is a value the limit takes | “The limit is $\infty$, so it exists” | $\infty$ is not a real number; writing $\lim = \infty$ describes how the limit fails |
| Checking only one sequence in oscillation | Evaluating $\sin(1/x)$ at $x = 1/(n\pi)$ and getting 0 each time, concluding limit is 0 | A limit requires every sequence to converge; a single convergent subsequence is not enough |
| Confusing jump with oscillation | “The function oscillates between 1 and $-1$” for the floor function at an integer | The floor function has a simple jump; oscillation means infinitely many fluctuations near the point |
Leveled Practice
Level 1 -- Classify the Failure
Problem 1. For each function, determine whether $\lim_{x \to 0} f(x)$ exists and, if not, classify the failure.
(a) $f(x) = \dfrac{x}{|x|}$
(b) $f(x) = \dfrac{1}{x}$
(c) $f(x) = \cos\!\left(\dfrac{1}{x}\right)$
Show answer
(a) Left limit $= -1$, right limit $= 1$. Jump failure. Limit does not exist.
(b) Left limit $= -\infty$, right limit $= +\infty$. Unbounded failure. Limit does not exist.
(c) As $x \to 0$, $\cos(1/x)$ oscillates between $-1$ and $1$ infinitely often. Oscillation failure. Limit does not exist.
Problem 2. Let $f(x) = \begin{cases} x + 1 & x < 2 \\ x^2 - 1 & x \geq 2 \end{cases}$.
Does $\lim_{x \to 2} f(x)$ exist? If not, classify the failure.
Show answer
Left limit: $2 + 1 = 3$. Right limit: $4 - 1 = 3$. Both equal 3. Limit exists and equals 3.
(Not a failure -- included to reinforce that not every piecewise boundary is a jump.)
Level 2 -- Deeper Analysis
Problem 3. Show that $\lim_{x \to 0} \sin(\pi / x)$ does not exist by finding two sequences $x_n \to 0$ that give different limits for $\sin(\pi/x_n)$.
Show answer
Let $x_n = 1/n$. Then $\sin(\pi/x_n) = \sin(n\pi) = 0$ for every positive integer $n$. This subsequence gives limit 0.
Let $y_n = \frac{2}{4n+1}$. Then $\pi/y_n = \frac{\pi(4n+1)}{2}$, and $\sin(\pi/y_n) = \sin\!\left(2n\pi + \frac{\pi}{2}\right) = \sin(\pi/2) = 1$. This subsequence gives limit 1.
Since two sequences approaching 0 give different values (0 and 1), no single limit exists. Oscillation failure.
Problem 4. The function $g(x) = \dfrac{\sin x}{x}$ at $x = 0$: explain why this is not one of the three failure modes, even though $g$ is not defined at $x = 0$.
Show answer
The limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ exists. Being undefined at $x = 0$ does not mean the limit fails -- the limit depends only on nearby values, not on $g(0)$. This is a removable discontinuity: the limit exists, but $g(0)$ is missing (or could be defined to equal 1 to make $g$ continuous).
Level 3 -- Construction
Problem 5. Construct a function $f$ that has a jump discontinuity at $x = 1$, a vertical asymptote at $x = 3$, and is continuous everywhere else on $[0, 5]$.
Show answer
One construction: $f(x) = \begin{cases} 0 & 0 \leq x < 1 \\ 2 & 1 \leq x < 3 \\ \frac{1}{(x-3)^2} & 3 < x \leq 5 \end{cases}$.
At $x = 1$: left limit $= 0$, right limit $= 2$, jump.
At $x = 3$: $f$ is undefined, $\frac{1}{(x-3)^2} \to +\infty$ from both sides, vertical asymptote.
On $[0,1)$, $(1,3)$, and $(3,5]$: each piece is continuous.
(Many other constructions work; what matters is the features, not the specific formulas.)
Common Misconceptions
a limit always fails because the function is not defined at the point. Students sometimes think the only reason a limit fails is that the function has no value at $a$. But the three failure modes have nothing to do with whether $f(a)$ exists. A jump discontinuity can occur even when $f(a)$ is defined (one of the two side values is chosen). Oscillation near a point, as with $\sin(1/x)$ near 0, fails the limit even though $f$ is defined for all $x \neq 0$. The function being defined or undefined at $a$ is irrelevant to whether the limit exists.
a limit not existing means the function is discontinuous. A function can fail to have a limit at a point and still be perfectly well-behaved everywhere else. Failure of a limit at one point says nothing about continuity at other points. Conversely, removing a jump discontinuity by redefining $f$ at exactly one point does not make the limit exist; the limit is controlled by the nearby behavior, not by the single point.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Think of the limit as the consensus a committee of observers must reach. Each observer approaches $x = a$ along a different path and reports what value $f(x)$ seems to be heading toward.
- Jump: two observers -- one from each side -- report different finite values. No consensus; limit fails.
- Vertical asymptote: observers report that $f(x)$ grows without bound. No finite consensus; limit fails.
- Oscillation: observers along different paths report wildly different values as they get closer to $a$. No consensus, not even close to one; limit fails.
A limit exists when every observer -- no matter how they approach $a$, from any direction, along any sequence -- agrees on the same finite value.
Connections
Within Calculus I (MATH161)
- Continuity: Continuous functions at $a$ never exhibit any of the three failures. Every failure mode corresponds to a type of discontinuity: jump discontinuity, infinite discontinuity, or oscillatory discontinuity.
- Differentiability: Corners in a graph arise when the left and right derivatives disagree -- a “jump” at the level of the derivative, not the function itself. If the function has a jump in the first place, it cannot be differentiable.
Toward Later Courses
- Sequences and series (MATH162): The concept that “every sequence approaching $a$ must converge to $L$” for a limit to exist is the same idea used to define limits of sequences.
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