The Limit of a Function

The limit of $f(x)$ as $x$ approaches $a$ is the single value $f(x)$ gets arbitrarily close to as $x$ nears $a$ from either side. It's about the journey toward $a$, never the arrival — so a limit can exist even where $f(a)$ is undefined or differs from it.

By the end you'll be able to read $\lim_{x \to a} f(x)$ off a graph or a table of values, decide whether a two-sided limit exists by comparing the left-hand and right-hand approaches, and explain why a hole in the graph doesn't stop a limit from existing.

Predict: as $x$ slides toward 2 from both sides, does $f(x)$ settle on a single number even though there's a hole exactly at $x = 2$? Drag the slider to check.

This is $f(x) = \dfrac{x^2 - 4}{x - 2}$ — algebraically the line $x + 2$, except $f(2)$ itself is $0/0$: undefined, an open circle. The dashed target line marks the value the outputs crowd toward. Then hit the toggle to see a jump function where the two sides head to different values — and the two-sided limit dies.

x = 0.500

x = 0.500 · f(x) = 2.500 · distance from a = 2: 1.500

f(x) = (x² − 4)/(x − 2) — a hole at x = 2, yet both sides crowd toward 4

x	1.9	1.99	1.999	→ 2 ←	2.001	2.01	2.1
f(x)	3.9	3.99	3.999	→ 4 ←	4.001	4.01	4.1

One definition, two tests: a limit is the single value the outputs approach, and it exists precisely when the approach from the left agrees with the approach from the right — regardless of what happens (or doesn't) at the point itself.

Intuitive

Writing $\lim_{x \to a} f(x) = L$ means: you can force $f(x)$ to lie as near $L$ as you please by keeping $x$ sufficiently near $a$ — but not equal to it. The point $x = a$ is deliberately excluded from the discussion. That's the whole trick you saw in the demo: the hole sits exactly at $x = 2$, yet the outputs crowd toward 4 anyway, because the limit only cares about the journey toward $a$, never the value at $a$. This is why a limit can exist where $f(a)$ is undefined, and can disagree with $f(a)$ when the graph has a misplaced point.

Formal

Split the approach by direction. The left-hand limit $\lim_{x \to a^-} f(x)$ looks only at $x < a$; the right-hand limit $\lim_{x \to a^+} f(x)$ looks only at $x > a$. The two-sided limit exists if and only if both one-sided limits exist and are equal: $$\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L = \lim_{x \to a^+} f(x).$$ If the two sides disagree — as at the jump in the demo, where the left heads to 5 and the right to 1 — the two-sided limit does not exist.

Numerical

You can watch a limit form in a table. For $f(x) = \dfrac{x^2 - 1}{x - 1}$, substituting $x = 1$ gives $0/0$ — an indeterminate form — yet the pattern is unmistakable: at $x = 0.9, 0.99, 1.001, 1.01$ the outputs are $1.9, 1.99, 2.001, 2.01$. From both sides $f(x) \to 2$, so $\lim_{x \to 1} f(x) = 2$, even though $f(1)$ is undefined — the graph is the line $y = x + 1$ with a hole at $x = 1$.

Worked example

Evaluate $\lim_{x \to 1} \dfrac{x^2 - 1}{x - 1}$. Substitution gives $\tfrac{0}{0}$ — indeterminate, so factor instead: $\dfrac{(x-1)(x+1)}{x-1} = x + 1$ for all $x \ne 1$. Near $x = 1$ (but never at it), $f$ behaves exactly like $x + 1$, so $f(x) \to 1 + 1 = $ 2. The limit is 2; $f(1)$ remains undefined — and that's fine.

Your turn

Same idea, new numbers: evaluate $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$. Substitution gives $\tfrac{0}{0}$, so factor: $\dfrac{(x-3)(x+3)}{x-3} = x + 3$ for all $x \ne 3$. Now finish it: as $x \to 3$, $f(x) \to $ ____ — and is $f(3)$ defined?

Reveal the answer

$f(x) \to 3 + 3 = $ 6, so $\lim_{x \to 3} f(x) = 6$. And no — $f(3) = \tfrac{0}{0}$ is undefined. The limit exists anyway: the graph is the line $y = x + 3$ with a hole at $x = 3$, just like the demo's function had a hole at $x = 2$.

More info — why the hole can't stop the limit

Think about what the graph and the table are actually using: only inputs near $a$, never $a$ itself. Every row of the approach table — 1.9, 1.99, 1.999 from the left; 2.1, 2.01, 2.001 from the right — is a legal input where the function is perfectly defined, and their outputs squeeze toward one value. The single missing point can't interrupt that squeeze, because it was never consulted. Replay this with the slider in the demo above: you can get $x$ as close to 2 as the slider allows without ever landing on it, and $f(x)$ tracks toward 4 the whole way. The Math is Fun link in Dive deeper below walks the same hole-at-a-point idea through a gentle $0/0$ example.

Check your understanding

Question 1 of 4

Evaluate $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$. Is $f(2)$ defined?

Question 2 of 4

A function has $\lim_{x \to 3^-} f(x) = 5$ and $\lim_{x \to 3^+} f(x) = 1$. What is $\lim_{x \to 3} f(x)$?

Question 3 of 4

Suppose $\lim_{x \to a} f(x) = L$ exists. Which statement MUST be true?

Question 4 of 4

Interleaving this lesson's two sub-ideas — check the one-sided limits before deciding hole vs. jump. Define $k(x) = 3x - 1$ for $x < 6$, $k(6) = 20$, and $k(x) = 2x + 5$ for $x > 6$. What is $\lim_{x \to 6} k(x)$, and is this a hole or a jump?

Recap

$\lim_{x \to a} f(x) = L$ means $f(x)$ gets arbitrarily close to the single value $L$ as $x$ nears $a$ from both sides — with $x = a$ itself deliberately excluded.
The two-sided limit exists if and only if the left-hand limit and right-hand limit both exist and are equal; if they disagree (a jump), it does not exist.
A limit can exist even where $f(a)$ is undefined (a hole), and can differ from $f(a)$ (a misplaced point) — the limit is about the approach, not the value at the point.
$0/0$ from substitution is an indeterminate form — a signal to do more work (factor, tabulate, graph), not a verdict that the limit fails to exist.

Dive deeper

OpenStax Calculus Volume 1 — 2.2 The Limit of a Function Read the intuitive and numerical build-up to limits with tables, graphs, and one-sided limits
Paul's Online Notes — The Limit Reinforce the arbitrarily-close intuition with numerical tables
Math is Fun — Limits (An Introduction) Preview the hole-at-a-point idea with a gentle 0/0 example

Sources