How is the limit of a function at a point different from its value there?

The value is a number the function actually takes at that argument — and it may not exist at all. The limit instead describes where the values are heading for arguments close to the point but different from it. That is why (x² − 4)/(x − 2) has no value at x = 2 and still has a limit of 4 there.

When can a limit be computed by plain substitution?

When the function is continuous at that point — and polynomials are continuous everywhere, rational functions everywhere except the zeros of the denominator. For those the limit simply is the value, so you put the number into the formula. Substitution only fails when it lands on an indeterminate form.

What do I do when substitution gives 0/0?

It signals that the numerator and the denominator share a factor — the one that makes both vanish. Factor them, cancel it, and substitute only then. The symbol 0/0 is not a result but a message that the computation has to be done differently: the same 0/0 can end up as 4, as 0, or as anything else.

How do you take a limit at infinity of a rational function?

Compare the degrees of the numerator and the denominator. With equal degrees the limit is the ratio of the leading coefficients, so (3x² + 2)/(x² − 1) tends to 3. When the numerator has the lower degree, the denominator wins the race and the limit is 0. When it has the higher degree, the function grows without bound and no finite limit exists.

Can a limit be infinite?

Then we speak of an improper limit: the values close in on no number at all, they simply grow (or fall) without bound. Writing lim(x → 0⁺) 1/x = +∞ does not say that the limit is a number called ∞ — it is shorthand for a behaviour. That is why the exercises in this lesson ask only about finite limits.

When does a limit fail to exist?

In three typical situations: when the one-sided limits differ (a function that jumps), when the function escapes to +∞ on one side and to −∞ on the other (as 1/x does at zero), and when the values oscillate without settling on a direction. In each case the correct answer is the sentence "the limit does not exist", not a number.

Why compute limits at all?

Because without them there is no derivative and no integral. A derivative is the limit of a difference quotient — an expression of exactly the 0/0 kind, the one substitution cannot handle. The whole of calculus rests on this single idea, and outside mathematics a limit describes any situation phrased as "as we get closer and closer to".

Advanced level

Limit of a function

Q: What are one-sided limits?

They are limits taken from one side only: the left-hand limit uses arguments smaller than x₀, the right-hand one arguments larger than it. The ordinary limit exists exactly when both one-sided limits exist and are equal. If they differ, there is no limit at that point — even though each side on its own is perfectly well defined.

A limit says where the values of a function are heading as the argument closes in on a number — even when the function has no value at that number at all. It is the first idea of analysis and the ground the derivative stands on.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Reading properties from a graphA graph shows the whole function at once: domain and range, zeros, sign, intervals of monotonicity and the extremum. Learn to read them off the drawing before you compute anything.
Quadratic equationsA quadratic equation has the unknown in the second power. Learn the general form, the discriminant, the formulas for the roots, the shortcuts for incomplete equations, Vieta formulas and how to read the solutions off a parabola.

All formulas

Limit of a function
$\lim_{x \to x_0} f(x) = g$
the values f(x) close in on g as x closes in on x₀
Continuous function
$\lim_{x \to x_0} f(x) = f(x_0)$
a polynomial is continuous, so the limit is plain substitution
The 0/0 indeterminate form
$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4$
cancel the common factor first, substitute afterwards
Equal degrees
$\lim_{x \to \infty} \frac{a_n x^n + \ldots}{b_n x^n + \ldots} = \frac{a_n}{b_n}$
the ratio of the leading coefficients
Numerator of lower degree
$\lim_{x \to \infty} \frac{a_m x^m + \ldots}{b_n x^n + \ldots} = 0 \quad (m < n)$
the denominator grows faster, so the quotient dies away
One-sided limits
$\lim_{x \to x_0^-} f(x) = \lim_{x \to x_0^+} f(x) = g \iff \lim_{x \to x_0} f(x) = g$
a limit exists only when both sides agree

Quantities rarely interest us at a single point; what matters is the approach to it. That is what a limit describes: which number the values $f(x)$ are heading for as the argument closes in on $x_0$ — while the point $x_0$ itself is never taken into account.

\lim_{x \to x_0} f(x) = g

Read it as "the limit of $f$ as $x$ tends to $x_0$ equals $g$ ". Notice that the sentence never says "value" — and that is no accident.

When substitution is enough

For a continuous function — and a polynomial is continuous at every point — the limit simply is the value:

\lim_{x \to x_0} f(x) = f(x_0)

So $\lim_{x \to 3} (x^2 - 2x) = 9 - 6 = 3$ , with no trickery involved: the graph has neither a hole nor a jump there, so "heading towards" and "being at" agree. Most limits you will meet work this way, which is why the first move is always to try substituting.

The trouble starts only when substitution returns nonsense.

The 0/0 indeterminate form

Look at the function

f(x) = \frac{x^2 - 4}{x - 2}

At $x = 2$ the denominator is zero, so the function has no value there — its domain is every number except two. Substituting gives $\tfrac{0}{0}$ , which is nothing at all. And yet the graph looks like this:

Everywhere except x = 2 our function is the plain line y = x + 2. At x = 2 the graph has a one-point hole — and that is precisely where we ask for the limit.

Because for every $x \neq 2$ the shared factor may be cancelled:

\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2

A limit never looks inside the point $x_0$ , so the cancelling is entirely legitimate:

\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4

The symbol $\tfrac{0}{0}$ is called indeterminate for exactly this reason: on its own it means nothing. Two other functions landing on $\tfrac{0}{0}$ at a point may have limit $0$ , limit $-7$ , or none. It is not a result but a note that the computation has to go another way.

Compute lim(x → 3) (x² − 9)/(x − 3) and lim(x → 1) (x² − 5x + 4)/(x − 1).

Limits at infinity

The second question we put to a function is what happens to it as the argument grows without end. For a rational function the answer is decided by comparing degrees.

The curve never touches the line y = 3, yet it gets arbitrarily close to it. Such a line is called a horizontal asymptote.

With equal degrees only the ratio of the leading coefficients counts — the remaining terms grow too slowly to change anything:

\lim_{x \to \infty} \frac{3x^2 + 2}{x^2 - 1} = \frac{3}{1} = 3

When the numerator's degree is lower, the denominator grows faster and the quotient dies away:

\lim_{x \to \infty} \frac{2x + 5}{x^2 + 1} = 0

Formally both cases are one move: divide the numerator and the denominator by the highest power of the denominator and use the fact that $\tfrac{1}{x} \to 0$ .

Compute lim(x → ∞) (6x² − x)/(2x² + 7) and lim(x → ∞) (4x + 1)/(x² − 3).

One-sided limits

A point can be approached from two sides — and sometimes that makes a difference. The left-hand limit takes only arguments smaller than $x_0$ , the right-hand limit only larger ones:

\lim_{x \to x_0^-} f(x), \qquad \lim_{x \to x_0^+} f(x)

The ordinary limit exists exactly when both one-sided limits exist and agree:

\lim_{x \to x_0^-} f(x) = \lim_{x \to x_0^+} f(x) = g \iff \lim_{x \to x_0} f(x) = g

When there is no limit

The cleanest example is $f(x) = \tfrac{1}{x}$ at zero:

The two branches of the hyperbola 1/x. Approaching zero from the right the values grow without end; from the left they fall without end. There is no common limit.

From the right the values grow without bound, from the left they fall without bound:

\lim_{x \to 0^+} \frac{1}{x} = +\infty, \qquad \lim_{x \to 0^-} \frac{1}{x} = -\infty

Limits like these are called improper. Writing $\infty$ does not claim that the limit is some number — it is shorthand for "grows without bound". And since the two sides escape in opposite directions, the limit $\lim_{x \to 0} \tfrac{1}{x}$ does not exist.

Three typical situations in which there is no limit:

different one-sided limits — the function jumps, the way a tariff changes at midnight;
escape to opposite infinities — the case of $\tfrac{1}{x}$ above;
oscillation — the values keep swinging and never settle on a direction.

A car park charges £5 for the first hour and £9 for every further hour started. Does the cost have a limit as the time tends to 1 hour?

What limits are for

They are the only tool that answers the question "how big is the change at a point". The difference quotient — the expression the derivative is about to be built from — heads towards $\tfrac{0}{0}$ as the gap shrinks, so without a limit it cannot be evaluated at all.

The same mechanism lets the region under a curve be sliced into ever thinner strips and pushed to a limit, at which moment the sum of rectangle areas becomes an integral. Both of the lessons that follow rest on this one idea.

Exercises

The exercises cover the three situations of this lesson: the limit of a polynomial (plain substitution), the $\tfrac{0}{0}$ form that needs cancelling, and a limit at infinity. All of them are finite, so the answer is always a number — the cases with $\infty$ and "no limit" are settled by reasoning, not by typing.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

lim(x → 4) (3x² + 5x + 9) =

Common mistakes

Treating 0/0 as a result — it is an indeterminate form, an instruction to compute differently, not a number.
Confusing the limit with the value — a function may have no value at a point and still have a limit there.
Substituting ∞ into the formula — infinity is not a number; compare degrees or divide by the highest power.
Declaring a limit to exist after checking one side — both one-sided limits must agree.
Cancelling without factoring — in $\tfrac{x^2 - 4}{x - 2}$ you may not cancel the bare "x"; factor first, cancel second.
Writing an improper limit as a number — $+\infty$ describes a behaviour, it is not an answer to type in.

Formula card

Topic: Limit of a function

Limit of a function
$\lim_{x \to x_0} f(x) = g$
the values f(x) close in on g as x closes in on x₀
Continuous function
$\lim_{x \to x_0} f(x) = f(x_0)$
a polynomial is continuous, so the limit is plain substitution
The 0/0 indeterminate form
$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4$
cancel the common factor first, substitute afterwards
Equal degrees
$\lim_{x \to \infty} \frac{a_n x^n + \ldots}{b_n x^n + \ldots} = \frac{a_n}{b_n}$
the ratio of the leading coefficients
Numerator of lower degree
$\lim_{x \to \infty} \frac{a_m x^m + \ldots}{b_n x^n + \ldots} = 0 \quad (m < n)$
the denominator grows faster, so the quotient dies away
One-sided limits
$\lim_{x \to x_0^-} f(x) = \lim_{x \to x_0^+} f(x) = g \iff \lim_{x \to x_0} f(x) = g$
a limit exists only when both sides agree

After cancelling, (x² − 4)/(x − 2) is the same line y = x + 2 everywhere except at x = 2. The function has no value there, yet the limit is 4.

For large x the graph of f(x) = (3x² + 2)/(x² − 1) hugs the horizontal line y = 3 — that line is the limit at infinity.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min