If I get $\frac{0}{0}$, does the limit exist?

You cannot tell yet. $\frac{0}{0}$ can hide any value: $\lim_{x \to 0} \frac{2x}{x} = 2$, $\lim_{x \to 0} \frac{x^2}{x} = 0$. Simplify first, then decide.

When should I rationalize instead of factoring?

When the 0 comes from a difference with a square root, like $\sqrt{x} - 2$ or $\sqrt{x + 4} - 2$. Multiply top and bottom by the conjugate (same terms, opposite sign in the middle), because $(\sqrt{x} - 2)(\sqrt{x} + 2) = x - 4$ has no root left.

Is the limit the same as the function value?

For polynomials and for rational functions where the denominator is not 0, yes: that is exactly what direct substitution says. At a hole, the function value does not exist, but the limit does (lessons Continuity and Types of Discontinuity go deeper).

Teacher Resources

Limits by Direct Substitution

Q: Why may I cancel $(x - 3)$ if it is 0 at $x = 3$?

A limit as $x \to 3$ only uses $x$ values close to 3, never $x = 3$ itself. For all those values $x - 3 \neq 0$, so cancelling is a legal step. The two expressions agree everywhere except at $x = 3$, and the limit does not care about that single point.

Find limits algebraically: plug in the value when you can, and when you get 0/0, factor and cancel or rationalize first.

Advanced30 minLesson

Printable worksheet

Definition

Direct substitution means finding a limit by simply plugging the value into the function:

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

This works for every polynomial and for every rational function whose denominator is not zero at

x = a

. Their graphs have no gaps there, so the value the function approaches is the value it actually takes.

Why it works: the limit laws. If

\lim_{x \to a} f(x)

and

\lim_{x \to a} g(x)

exist, then limits can be split up:

sum and difference: $\lim (f \pm g) = \lim f \pm \lim g$
constant multiple: $\lim (c \cdot f) = c \cdot \lim f$
product: $\lim (f \cdot g) = \lim f \cdot \lim g$
quotient: $\lim \frac{f}{g} = \frac{\lim f}{\lim g}$ , only if $\lim g \neq 0$
power: $\lim f^n = (\lim f)^n$

What the substitution tells you:

a number, like $\frac{5}{5} = 1$ : that number is the limit
$\frac{\text{nonzero}}{0}$ , like $\frac{3}{0}$ : no finite limit (the graph has a vertical asymptote, see the lesson Infinite Limits)
$\frac{0}{0}$ : the indeterminate form 0/0. It is NOT an answer. It tells you to rewrite the expression first.

Fixing 0/0:

1.Factor and cancel: both parts share a factor $(x - a)$ . Cancel it, then substitute again.
2.Rationalize: with a square root like $\sqrt{x} - 2$ , multiply top and bottom by the conjugate $\sqrt{x} + 2$ , cancel, then substitute again.

Cancelling is allowed because the limit only looks at

x

values near

a

, never at

x = a

itself, so

x - a \neq 0

Try it now

Which limit can you find right away by direct substitution?

Worked Examples

Find $\lim_{x \to 3} (x^2 - 2x + 3)$ .

Check the type of function

$x^2 - 2x + 3$ is a polynomial, defined for every $x$ → Direct substitution works

Substitute $x = 3$

$3^2 - 2 \cdot 3 + 3 = 9 - 6 + 3$ → $6$

Write the limit

$\lim_{x \to 3} (x^2 - 2x + 3) = 6$ → The limit is $6$

Answer: $\lim_{x \to 3} (x^2 - 2x + 3) = 6$ . On the first graph the point $(3, 6)$ lies on the curve and the table values close in on $6$ .

Common Mistakes

Stopping at $\frac{0}{0}$ and answering "the limit is 0", "1" or "does not exist"

Why it's wrong: $\frac{0}{0}$ is not a number and not a final answer. It only says that the top and bottom share a factor that hides the real value.

Correct: Treat $\frac{0}{0}$ as a signal: factor and cancel (or rationalize), then substitute again. $\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3} = 5$ .

Cancelling terms instead of factors, like $\frac{x^2 - 9}{x - 3} = \frac{x - 9}{-3}$

Why it's wrong: You may only cancel a factor that multiplies the whole numerator and the whole denominator. $x$ inside a sum is not such a factor.

Correct: Factor first: $\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3$ , so the limit as $x \to 3$ is $6$ .

Treating $\frac{3}{0}$ like $\frac{0}{0}$ and trying to factor

Why it's wrong: When only the denominator is 0, there is no common factor to cancel. The values grow without bound near that $x$ .

Correct: $\frac{\text{nonzero}}{0}$ means there is no finite limit (vertical asymptote). Only $\frac{0}{0}$ calls for factoring or rationalizing.

Multiplying only the numerator by the conjugate

Why it's wrong: Multiplying just one part changes the value of the fraction. You are allowed to multiply by $\frac{\sqrt{x} + 2}{\sqrt{x} + 2}$ , which equals 1.

Correct: Multiply top AND bottom by the conjugate, and leave the bottom factored so you can cancel $(x - 4)$ .

Interactive Visual

Function graph

Tap or hover over the graph to read off values.

f(x) = x² − 2x + 3

Table of values

x	2.9	2.99	3	3.01	3.1
y	5.61	5.9601	6	6.0401	6.41

Function graph

Tap or hover over the graph to read off values.

g(x) = (x² − x − 6)/(x − 3)

Table of values

x	2.9	2.99	3.01	3.1
y	4.9	4.99	5.01	5.1

Function graph

Tap or hover over the graph to read off values.

h(x) = (√x − 2)/(x − 4)

Table of values

x	3.9	3.99	4.01	4.1
y	0.251582	0.250156	0.249844	0.248457

Interactive Sandbox

Interactive Grapher

Tap or hover over the graph to read off values.

Try these examples:

y = 2x + 1

m=2, b=1

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Practice Problems

18 problems

Problem 1 of 18

Easy

Which limit can you find right away by direct substitution?

Why It Matters

In the lessons Evaluating Limits Graphically and Evaluating Limits from Tables you estimated limits by looking. Direct substitution gives the exact value in seconds, and it is the method you will use most often in calculus.

Derivatives: every derivative starts as a limit of the form $\frac{0}{0}$ , for example $\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}$ . Factoring and cancelling is exactly how you solve it.
Physics: instantaneous speed is the limit of average speeds, found by factoring a difference quotient.
Continuity: a function is continuous at $a$ exactly when direct substitution works there (lesson Continuity).
Holes in graphs: the factor you cancel shows where a rational function has a hole, and the limit tells you its height.

Knowing when you may substitute, and what to do when you get

\frac{0}{0}

, saves you from wrong answers like "the limit is 0" or "the limit does not exist".

Real World Applications

Instantaneous Speed

A speedometer shows speed at one instant. Physicists find it as the limit of average speeds over shorter and shorter time intervals, and that limit always starts as 0/0.

Example:

A cart moves so that its position is $s(t) = t^2 + 3t$ meters after $t$ seconds. Its average speed between second 2 and time $t$ is $\frac{s(t) - s(2)}{t - 2} = \frac{t^2 + 3t - 10}{t - 2}$ . Factoring gives $\frac{(t - 2)(t + 5)}{t - 2} = t + 5$ .

1Try It Yourself

The cart's position is $s(t) = t^2 + 3t$ meters. Its speed at $t = 2$ seconds is $\lim_{t \to 2} \frac{t^2 + 3t - 10}{t - 2}$ .

What is the cart's speed at $t = 2$ seconds, in meters per second?

Step 1: Write the mathematical expression

Factor, cancel and substitute: the speed is

Growing Square Tiles

Designers and engineers ask how fast an area changes when a length changes. For a square of side $x$ the area is $x^2$, and the rate of change at one size is a 0/0 limit.

Example:

A square tile with side $x$ cm has area $A(x) = x^2$ cm². When the side grows from 5 cm to $x$ cm, the area changes on average by $\frac{x^2 - 25}{x - 5}$ cm² per cm of side.

2Try It Yourself

The rate at which the tile's area grows when the side is exactly 5 cm is $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$ .

How many cm² per cm does the area grow at a side of 5 cm?

Step 1: Write the mathematical expression

Factor $x^2 - 25$ , cancel and substitute: the rate is

Plant Growth Rate

Biologists model growth with square root formulas. The growth rate on a given day is a limit that needs rationalizing.

Example:

A seedling is $h(t) = \sqrt{t}$ dm tall after $t$ days. Its average growth between day 9 and day $t$ is $\frac{\sqrt{t} - 3}{t - 9}$ dm per day.

3Try It Yourself

The growth rate on day 9 is $\lim_{t \to 9} \frac{\sqrt{t} - 3}{t - 9}$ . Multiplying top and bottom by $\sqrt{t} + 3$ turns it into $\lim_{t \to 9} \frac{1}{\sqrt{t} + 3}$ .

What is the growth rate on day 9, in dm per day?

Step 1: Write the mathematical expression

Substitute into the simplified expression: the rate is

Key Takeaways

1For polynomials, $\lim_{x \to a} f(x) = f(a)$ : just substitute
2For rational functions, substitute when the denominator is not 0 at $x = a$
3The limit laws let you split a limit into sums, products, quotients (denominator limit not 0) and powers
4 $\frac{0}{0}$ is an indeterminate form, never the answer: rewrite the expression first
5Factor and cancel the common factor $(x - a)$ , or multiply by the conjugate when there is a square root, then substitute again
6 $\frac{\text{nonzero}}{0}$ means no finite limit (vertical asymptote), not a hole

Frequently Asked Questions

A limit as

x \to 3

only uses

x

values close to 3, never

x = 3

itself. For all those values

x - 3 \neq 0

, so cancelling is a legal step. The two expressions agree everywhere except at

x = 3

, and the limit does not care about that single point.

Glossary

Direct substitution: Finding $\lim_{x \to a} f(x)$ by computing $f(a)$ ; works for polynomials and for rational functions with a nonzero denominator at $a$
Limit laws: Rules that split a limit of a sum, difference, product, quotient or power into limits of its parts
Indeterminate form 0/0: The result $\frac{0}{0}$ after substitution; it does not give a value and means the expression must be rewritten
Conjugate: For $\sqrt{x} - b$ the conjugate is $\sqrt{x} + b$ ; their product $x - b^2$ has no square root
Rationalizing: Multiplying top and bottom by a conjugate to remove a square root from a difference

Formula Card

Direct substitution

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

For polynomials, and for rational functions when the denominator is not 0 at $a$

Quotient law

\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}

Only when $\lim_{x \to a} g(x) \neq 0$

Factor and cancel

\frac{(x - a) \cdot p(x)}{(x - a) \cdot q(x)} = \frac{p(x)}{q(x)}

Use when substitution gives $\frac{0}{0}$, then substitute again

Conjugate product

(\sqrt{x} - b)(\sqrt{x} + b) = x - b^2

Use to rationalize a 0/0 limit with a square root

Limits by Direct Substitution

Definition

Worked Examples

Common Mistakes

Interactive Visual

Function graph

Table of values

Function graph

Table of values

Function graph

Table of values

Interactive Sandbox

Interactive Grapher

Expression Calculator

History

Practice Problems

Why It Matters

Real World Applications

Instantaneous Speed

Growing Square Tiles

Plant Growth Rate

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Direct substitution

Quotient law

Factor and cancel

Conjugate product

More in This Topic

Introduction to Limits

Limit Notation

Evaluating Limits Graphically

Evaluating Limits from Tables