Topic 1 of 7

Limits and continuity

Step 15 · Calculus 1 topics

Find limits from tables, graphs and algebra, use the limit laws, and decide where a function is continuous.

What a limit means

The limit of f(x) as x approaches a is the value f(x) gets close to as x gets close to a, from both sides. What happens at a itself does not matter: f(a) can be different, or missing.

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

y = (x² − 1) ÷ (x − 1) is the line y = x + 1 with a hole at x = 1. As x approaches 1, y approaches 2.

Example: From a table

Estimate $\lim_{x \to 0} \frac{\sin x}{x}$ , with x in radians.

x = 0.1 gives 0.99833, and x = 0.01 gives 0.99998.
Negative x give the same values, because sin x ÷ x is even.

Answer1

Example: One-sided limits

f(x) is x + 1 for x < 2 and x² for x ≥ 2. Does the limit as x approaches 2 exist?

From the left, x + 1 approaches 3.
From the right, x² approaches 4.
The two sides disagree.

AnswerNo: the graph jumps at x = 2

Limit laws and algebra

Limits respect arithmetic: the limit of a sum is the sum of the limits, and the same goes for differences, products, quotients (when the bottom's limit is not 0) and powers. So for polynomials and other well-behaved functions you can just substitute.

Example: Substitute

Find $\lim_{x \to 3} (x^{2} - 2x + 5)$ .

It is a polynomial, so put x = 3 in.
9 − 6 + 5 = 8.

Answer8

Example: Factor and cancel

Find $\lim_{x \to 3} \frac{x^{2} - 9}{x - 3}$ .

Substituting gives 0 ÷ 0, which says nothing yet.
Factor: (x − 3)(x + 3) ÷ (x − 3) is x + 3 for x ≠ 3.
Now substitute: 3 + 3 = 6.

Answer6

Example: Rationalize

Find $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ .

Multiply the top and bottom by $\sqrt{x + 4} + 2$ .
The top becomes (x + 4) − 4, which is x, so the fraction is $\frac{1}{\sqrt{x + 4} + 2}$ .
Substitute x = 0 to get 1 ÷ 4.

Answer0.25

Continuity

f is continuous at a when three things hold: f(a) is defined, the limit as x approaches a exists, and the two are equal. Informally, you can draw the graph through a without lifting your pen.

Break	What fails	Example
Hole	f(a) missing or wrong	(x² − 1) ÷ (x − 1) at 1
Jump	Left and right limits differ	The example above at 2
Asymptote	f(x) blows up	1 ÷ x at 0

Example: Fixing a hole

f(x) = (x² − 4) ÷ (x − 2) for x ≠ 2. What value of f(2) makes f continuous?

For x ≠ 2, f(x) is x + 2.
Its limit as x approaches 2 is 4.

Answerf(2) = 4

Example: Intermediate values

Does x³ + x − 1 = 0 have a solution between 0 and 1?

The left side is a polynomial, so it is continuous.
At 0 it is −1, and at 1 it is 1.
A continuous function going from −1 to 1 must pass through 0.

AnswerYes, by the intermediate value theorem

Try it yourself

Work each one on paper first, then open the answer to check.

Find the limit of 4x − 7 as x approaches 3.
Show answer
5
Find $\lim_{x \to 5} \frac{x^{2} - 25}{x - 5}$ .
Show answer
10
Find $\lim_{x \to 2} \frac{x^{2} - 3x + 2}{x - 2}$ .
Show answer
1
As x approaches 4, f(x) approaches 7 from both sides, but f(4) = 2. What is the limit as x approaches 4?
Show answer
7
Is f(x) = 1 ÷ (x − 3) continuous at x = 3?
Show answer
No: it is not defined there, and it blows up.
Find $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$ .
Show answer
$\frac{1}{6}$