Math · Calculus I · Concept

Limits of functions and holes in graphs

A limit is the value $f (x)$ gets close to as $x$ gets close to a number $a$ , from both sides, whether or not $f (a)$ exists. Substitute $a$ when the function is defined there. When substitution gives $0 / 0$ , rewrite the expression first and then substitute.

A concrete case

Look near the point

Take $f (x) = (x^{2} - 4) / (x - 2)$ and ask what happens near $x = 2$ . At $x = 2$ itself the formula gives $(4 - 4) / (2 - 2) = 0 / 0$ , which is not a number. The function has no value there.

Now try inputs close to $2$ . At $x = 1.9$ the formula gives $(3.61 - 4) / (- 0.1) = 3.9$ . At $1.99$ it gives $3.99$ , and at $2.01$ it gives $4.01$ . From both sides, the outputs close in on $4$ .

That $4$ is the limit of $f (x)$ as $x$ approaches $2$ . The function never takes the value $4$ , because it has no value at $2$ . A limit describes where the outputs are heading, not where they arrive.

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

$f (x)$ $= (x^{2} - 4) / (x - 2)$ near $x = 2$
$x$	$f (x)$
$1.9$	$3.9$
$1.99$	$3.99$
$2$	undefined $(0 / 0)$
$2.01$	$4.01$
$2.1$	$4.1$

The rule

What the limit notation says

The statement $lim f (x) = L$ as $x \to a$ means that you can make $f (x)$ as close to $L$ as you like by taking $x$ close enough to $a$ , with $x \neq a$ . Here $a$ is the input you approach, and $L$ is the output the values approach. Aloud, it reads “the limit of $f (x)$ as $x$ approaches $a$ is $L$ .”

The value $f (a)$ plays no part. $f$ may be undefined at $a$ , as in the case above, or it may have some other value there. Only the inputs near $a$ count.

You can also approach from one side. The left-hand limit, written with $x \to a^{-}$ , uses only inputs below $a$ . The right-hand limit, written with $x \to a^{+}$ , uses only inputs above it. The two-sided limit exists exactly when both one-sided limits exist and are equal.

lim_{x \to a} f (x) = L lim_{x \to a^{-}} f (x) = L = lim_{x \to a^{+}} f (x)

\begin{array}{c} lim_{x \to a} f (x) = L \\ lim_{x \to a^{-}} f (x) = L = lim_{x \to a^{+}} f (x) \end{array}

When substitution is enough

Limits combine the way arithmetic does. The limit of a sum is the sum of the limits, and the same holds for differences, products and constant multiples. It holds for quotients too, as long as the limit of the denominator is not $0$ .

For polynomials, roots, exponentials, logarithms and the trig functions, the limit at a point where the function is defined is the value there. That gives the first thing to try, direct substitution. The limit of $x^{2} + 1$ as $x$ approaches $3$ is $3^{2} + 1 = 10$ .

Substitution fails only when it produces something that is not a number, such as $0 / 0$ or $5 / 0$ . Those are the limits that need work.

lim_{x \to 3} (x^{2} + 1) = 3^{2} + 1 = 10

Why it works

Why canceling gives the right limit

For every $x$ except $2$ , the factor $x - 2$ is not $0$ , and it cancels. On all those inputs $(x^{2} - 4) / (x - 2)$ is exactly $x + 2$ . The two functions differ only at $x = 2$ , where the quotient is undefined and $x + 2$ equals $4$ .

A limit never looks at $x = 2$ itself. It looks only at inputs near $2$ , and there the two functions are identical. Functions that agree near a point have the same limit there. The limit of the quotient is therefore the limit of $x + 2$ , which substitution gives as $4$ .

This is also why $0 / 0$ is not an answer. It says only that the top and the bottom both shrink toward $0$ . What matters is how they compare as they shrink. $(x - 2) / (x - 2)$ approaches $1$ , $(x^{2} - 4) / (x - 2)$ approaches $4$ , and ${(x - 2)}^{2} / (x - 2)$ approaches $0$ , though all three give $0 / 0$ at $x = 2$ . Such a form is called indeterminate.

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

\begin{aligned} \frac{x^{2} - 4}{x - 2} & = \frac{(x - 2) (x + 2)}{x - 2} \\ = x + 2 (x \neq 2) \end{aligned}

The method

The method in four steps

Most limits in a first course fall to these four steps. The table works through the limit of $(x^{2} - 4) / (x - 2)$ as $x$ approaches $2$ .

Keep writing $lim$ in front of every row until the last one. The rows before it are limits, not values. Dropping $lim$ between rows turns a true statement into a false one, because $x + 2$ is not equal to $4$ .

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

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

Finding a limit by substitution, then by rewriting
Step	What to do	For $(x^{2} - 4) / (x - 2)$ at $x = 2$
1	Substitute $a$	$(4 - 4) / (2 - 2)$ $= 0 / 0$ , so keep going
2	A number is the limit. For $0 / 0$ , rewrite the expression	$(x - 2) (x + 2) / (x - 2)$
3	Cancel the common factor, which is allowed because $x \neq a$	$x + 2$ , for $x \neq 2$
4	Substitute $a$ into the simpler form	$2 + 2$ $= 4$

Other ways to open $0 / 0$

Factoring is one rewrite among several. Choose the one that fits the expression, and every one of them works the same way. It produces a function equal to the original for $x \neq a$ that you can substitute into.

For $(\sqrt{x} - 3) / (x - 9)$ at $x = 9$ , write $x - 9$ as $(\sqrt{x} - 3) (\sqrt{x} + 3)$ . The common factor cancels and leaves $1 / (\sqrt{x} + 3)$ , which gives $1 / 6$ .

Sines and cosines lean on one special limit, $sin u / u \to 1$ as $u \to 0$ , with $u$ in radians. Near $0$ the two are almost equal, and $sin (0.01) / 0.01 \approx 0.99998$ . On a circle of radius $1$ , $sin u$ is half the chord across an arc of length $2 u$ , and a short arc is almost as straight as its chord.

If substitution gives a nonzero number over $0$ , such as $5 / 0$ , no rewrite will help. The function grows without bound near $a$ , and the limit is not a number. Check each side. $1 / x^{2}$ heads to $\infty$ from both sides, while $1 / x$ heads to $- \infty$ from the left and to $\infty$ from the right.

Rewrites that open a $0 / 0$ limit
The expression has	Rewrite with	Example
polynomials on top and bottom	factoring and canceling	$(x^{2} - 4) / (x - 2)$ becomes $x + 2$
a square root	the conjugate, or a difference of squares	$(\sqrt{x} - 3) / (x - 9)$ becomes $1 / (\sqrt{x} + 3)$
a fraction inside a fraction	a common denominator	$(1 / x - 1 / 2) / (x - 2)$ becomes $- 1 / (2 x)$
sines and cosines	an identity and $sin u / u$ $\to 1$	$sin (5 x) / (3 x)$ becomes $(5 / 3) \cdot sin (5 x) / (5 x)$

Where it breaks

A limit can fail to exist in three ways. The two sides can head to different numbers. $∣ x ∣ / x$ is $- 1$ for every negative $x$ and $1$ for every positive $x$ . At $0$ the left-hand limit is $- 1$ and the right-hand limit is $1$ , so there is no two-sided limit.

The values can grow without bound. $1 / x^{2}$ is more than $10, 000$ once $x$ is within $0.01$ of $0$ . Writing the limit as $\infty$ records how it fails. $\infty$ is not a number the values approach.

The values can oscillate. $sin (π / x)$ swings between $- 1$ and $1$ faster and faster as $x$ approaches $0$ , and it approaches no single value.

That last case shows where a table misleads. At $x = 0.1$ , $0.01$ and $0.001$ , $sin (π / x)$ is exactly $0$ . At $x = 2 / 21$ and $2 / 201$ , which sit right beside those inputs, it is exactly $1$ . A table suggests a limit, and only algebra or the definition establishes one.

The opposite confusion is just as common. The limit at $a$ and the value at $a$ answer different questions. If $g (x) = x + 2$ for $x \neq 2$ and $g (2) = 7$ , the limit at $2$ is still $4$ .

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

\begin{array}{c} lim_{x \to 0^{-}} \frac{∣ x ∣}{x} = - 1 \\ lim_{x \to 0^{+}} \frac{∣ x ∣}{x} = 1 \end{array}

Check your answer

A limit found by rewriting can be tested without redoing the algebra.

Try numbers on both sides. For a limit at $2$ , work out $f (1.99)$ and $f (2.01)$ . Both should be close to your answer, and closer still at $1.999$ and $2.001$ .

Look at the graph. A $0 / 0$ that cancels shows as a hole, and the height of the hole is the limit.

Undo the rewrite. Multiply your simplified form by the factor you canceled, and you should get back the original numerator.

To have Chalk check the work, write it in a Math box with $lim$ in front of every row except the last. Put the limit you were given on the first row, each rewrite on its own row, and the value on the last. The Checker compares each row with the row above. A wrong value reads ✗ with the message “Not the limit in row 2.” If you drop $lim$ and write $x + 2$ above $4$ , the $4$ reads ✗, because $x + 2$ and $4$ are not equal.

A limit that does not exist can be written DNE on the row under the limit. It reads ✓ when the limit really has no value, such as $∣ x ∣ / x$ at $0$ , and ✗ when the limit is a number, such as $(x^{2} - 4) / (x - 2)$ at $2$ . When both sides run off to $\infty$ , as for $1 / x^{2}$ at $0$ , it reads ?, because courses write that case both ways. Explore in Limits compares the left and right sides.

The precise definition, and where limits lead

“As close as you like” has a precise form. For every tolerance $ε > 0$ there must be a distance $δ > 0$ with this property. Whenever $0 < ∣ x - a ∣ < δ$ , the output satisfies $∣ f (x) - L ∣ < ε$ . $ε$ says how close the outputs must be, and $δ$ says how close the inputs must be to guarantee it.

For the case above, $∣ f (x) - 4 ∣$ $= ∣ (x + 2) - 4 ∣$ $= ∣ x - 2 ∣$ whenever $x \neq 2$ . Taking $δ = ε$ works. Any input within $ε$ of $2$ , other than $2$ itself, gives an output within $ε$ of $4$ .

Limits are the foundation of the rest of calculus. Continuity asks whether the limit at a point equals the value there. The derivative is the limit of a $0 / 0$ form, the slope $(f (a + h) - f (a)) / h$ as $h$ approaches $0$ . The definite integral is a limit of sums.

0 < ∣ x - a ∣ < δ ⟹ ∣ f (x) - L ∣ < ε

Common mistakes

Calling $0 / 0$ the answer, or deciding from it that the limit does not exist. $0 / 0$ says only that the top and bottom both approach $0$ . Rewrite the expression and substitute again, and $(x^{2} - 4) / (x - 2)$ turns out to approach $4$ .
Dropping $lim$ between rows. Writing $(x^{2} - 4) / (x - 2)$ $= x + 2$ $= 4$ claims that $x + 2$ equals $4$ for every $x$ , which is false. Keep $lim$ on each row until you substitute.
Confusing the limit with the value. If $g (x) = x + 2$ near $2$ but $g (2) = 7$ , the limit at $2$ is $4$ , not $7$ . A limit ignores the value at the point itself.
Trusting a table as a proof. $sin (π / x)$ is $0$ at $x = 0.1$ , $0.01$ and $0.001$ , yet it has no limit at $0$ . A table suggests an answer, and the algebra establishes it.
Checking only one side. $∣ x ∣ / x$ approaches $1$ from the right of $0$ and $- 1$ from the left. The two-sided limit exists only when the left-hand and right-hand limits agree.

Key terms

Limit: The value $f (x)$ approaches as $x$ gets closer and closer to a point $a$ , not counting $x = a$ itself. A limit can exist even when $f (a)$ is undefined or has a different value.
One-sided limit: The limit as $x$ approaches $a$ from the left only $(x \to a^{-})$ or from the right only $(x \to a^{+})$ . The two-sided limit exists exactly when both one-sided limits exist and are equal.
Indeterminate form: A pattern such as $0 / 0$ , $\infty / \infty$ or $\infty - \infty$ that direct substitution produces and that doesn’t decide the limit on its own. Different limits can give the same pattern, so more work is needed, such as factoring or L’Hôpital’s rule.
Infinite limit: A statement that outputs eventually exceed every positive bound, or fall below every negative bound, as inputs approach the specified location. Infinity is not an ordinary finite output value.
Continuity: A function is continuous at a point when it is defined there and its limit there equals that value, so the graph has no hole, jump or break. On an interval this must hold at every point, using one-sided limits at closed ends.

Work through an example

Find the limit of $(x^{2} - 4) / (x - 2)$ as $x$ approaches $2$ .

Find a limit by factoring and canceling →

A trigonometric limit using $sin u / u \to 1$ →

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