Limits

Where a function is heading, even if it never arrives.

Take a graph with one point punched out of it. You can’t read the height at the hole, because there’s nothing there. But you can walk along the curve toward the hole from both sides and see exactly what height you’re heading for. That height is the limit, and it’s the idea the rest of calculus is built on.

On this page: the big idea, a graph explorer you can drag, one-sided limits, the three ways a limit can fail, limits at infinity, four worked examples, nine practice problems, and an eight-question quiz.

Try the explorer Jump to practice

The graph of sin x / x has a hole at x = 0, yet both sides head straight for the height 1.

Section IThe big idea: close, but not equal

Here’s a function that looks harmless:

f(x) = x² − 4x − 2

Try x = 2. The top is 4 − 4 = 0 and the bottom is 2 − 2 = 0, so you get 0/0. That isn’t a number. The function simply has no value at 2.

Now sneak up on 2 instead. At x = 1.9 you get 3.9. At 1.99 you get 3.99. At 2.01 you get 4.01. From both sides, the outputs are closing in on 4. We write:

limx → 2x² − 4x − 2 = 4

Read it aloud as “the limit, as x approaches 2, of f(x), is 4.”

Limit (informal definition). limx → af(x) = L means: you can make f(x) as close to L as you like by taking x close enough to a, from both sides, without x ever equaling a.

That last part is the heart of it. A limit describes the trip, not the destination. What happens exactly at x = a (a hole, a stray dot, or a perfectly normal value) has no effect on the limit.

Why care? Because the big questions in calculus, like “how fast is it changing at this instant?”, always lead to 0/0 if you plug in directly. Limits let you answer them anyway.

Section IILimit explorer

Pick a function, then drag the slider to move two points toward x = a: a red one from the left and a blue one from the right. The table fills in as you go. Watch what the outputs do.

Use the slider (or arrow keys) to shrink the distance between x and a from 1 down to 0.0001.

Function

How close? (distance from a)

Values from each side (red: left, blue: right)
x → a⁻	f(x)	x → a⁺	f(x)

This explorer needs JavaScript turned on. Here is what it shows. For f(x) = (x² − 4)/(x − 2): x = 1.9, 1.99, 1.999 give 3.9, 3.99, 3.999, and x = 2.1, 2.01, 2.001 give 4.1, 4.01, 4.001, so the limit at 2 is 4. For sin x / x: x = 0.1 gives 0.998334 and x = 0.01 gives 0.999983, so the limit at 0 is 1. For 1/x²: x = 0.01 gives 10,000, so the values blow up and the limit is ∞.

Section IIIOne-sided limits

Sometimes the left and right sides of a graph tell different stories. So we give each side its own notation:

limx → a⁻f(x) is the left-hand limit: x approaches a from below (values less than a).
limx → a⁺f(x) is the right-hand limit: x approaches a from above.

Take the piecewise function from the explorer: g(x) = x + 1 when x < 1, and g(x) = 4 − x when x ≥ 1.

Open dot: the graph doesn’t include that point. Filled dot: it does, so g(1) = 3.

Coming from the left, the heights head to 1 + 1 = 2. Coming from the right, they head to 4 − 1 = 3. So limx → 1⁻g(x) = 2 and limx → 1⁺g(x) = 3.

The two-sided rule. limx → af(x) = L exactly when both one-sided limits exist and both equal L.

Here the sides disagree (2 ≠ 3), so limx → 1g(x) does not exist, often shortened to DNE. Notice that g(1) = 3 exists just fine. A function can have a value at a point and still have no limit there.

Section IVThree ways a limit can fail

Most functions you meet have limits almost everywhere. When a limit doesn’t exist, it’s nearly always for one of these three reasons.

1. A jump

|x| / x is −1 for every negative x and 1 for every positive x. The sides disagree at 0.

2. A blow-up

1/x² grows past every number as x → 0. No finite limit; we write = ∞.

3. Endless wiggling

sin(1/x) swings between −1 and 1 infinitely often near 0, so it never settles on one height.

About “= ∞”: infinity isn’t a number, so writing limx → 01 over x² = ∞ is shorthand for “the limit does not exist, because the values grow without bound.” It describes how the limit fails.

Section VLimits at infinity

Limits can also ask about the far right end of a graph: what happens to f(x) as x gets enormous? If the outputs settle toward a number L, we write limx → ∞f(x) = L, and the line y = L is a horizontal asymptote.

For fractions of polynomials there’s a quick trick: divide every term by the highest power of x in the denominator. Any term like 2/x or 4/x² shrinks to 0 as x grows. Try the race below with

f(x) = 5x² + 2xx² + 4 = 5 + 2/x1 + 4/x² → 5 + 01 + 0 = 5

This graph needs JavaScript turned on. Sample values: f(1) = 1.4, f(10) = 5 exactly, f(20) ≈ 5.049505, f(1,000) ≈ 5.00198, f(1,000,000) ≈ 5.000002. The values settle toward 5.

Surprise: this graph crosses its asymptote. At x = 10, f(10) = 520/104 = 5 exactly, and just past that it rises a little above 5 before sinking back. That’s allowed. A horizontal asymptote describes the far-away behavior, not a line the graph can never touch.

Section VIWorked examples

Facing a limit, try the simplest move first: plug in. If you get a real number, you’re done. If you get 0/0, simplify and try again. If you get “nonzero ÷ 0”, look for a blow-up and check each side.

Example 1: direct substitution

Find limx → 3(x² + 2x − 1).

3² + 2(3) − 1Polynomials have no holes or jumps, so plug in= 9 + 6 − 1Simplify= 14Answer

✓ Numerical check: x = 2.999 gives 13.992001 and x = 3.001 gives 14.008001.

Example 2: factor away a hole

Find limx → 3x² − x − 6 over x − 3.

9 − 3 − 6 = 0, and 3 − 3 = 0Plugging in gives 0/0: there’s a hole to fix(x − 3)(x + 2)x − 3Factor the top: −3 · 2 = −6 and −3 + 2 = −1x + 2Cancel (x − 3), allowed because x ≠ 3 in a limit3 + 2 = 5Now plug in

The graph is the line y = x + 2 with one point missing at (3, 5).

✓ Numerical check: x = 2.999 gives 4.999 and x = 3.001 gives 5.001.

Example 3: one-sided limits that disagree

Find limx → 0|x| over x.

x < 0: |x| = −xFor negative x, absolute value flips the sign−xx = −1So the left side is always −1x > 0: |x| = xFor positive x, nothing changesxx = 1So the right side is always 1−1 ≠ 1: the limit does not existLeft-hand limit −1, right-hand limit 1

✓ Numerical check: x = −0.001 gives −1 and x = 0.001 gives 1. No matter how close you get, the sides stay 2 apart.

Example 4: a limit at infinity

Find limx → ∞3x² − 5x over 6x² + 1.

3 − 5/x6 + 1/x²Divide every term by x², the highest power in the bottom3 − 06 + 0As x grows, 5/x → 0 and 1/x² → 0= 12Answer

✓ Numerical check: x = 1,000 gives 0.499167; x = 1,000,000 gives 0.4999992.

Section VIICommon mistakes

“0/0 means the limit is 0” (or “doesn’t exist”). Neither. 0/0 is called an indeterminate form: it just means plugging in didn’t work yet. Simplify and try again.
“The limit is just f(a).” Only for well-behaved (continuous) functions like polynomials. Holes, jumps and stray points break this.
Checking only one side. A two-sided limit needs both sides to agree. Test numbers slightly below and slightly above.
Canceling without factoring. In x² − 4 over x − 2 you can’t cancel the 2s or the x’s. Factor first, then cancel whole factors.
Treating ∞ as a number. “∞ − ∞ = 0” and “∞/∞ = 1” are both false in general. Rewrite the expression, as in Example 4.
Degrees instead of radians. The fact that sin x / x → 1 only works with x in radians. Set your calculator to radian mode.

Section VIIIPractice problems

Try each one on paper first, then open the answer to check your steps. Every answer includes a quick numerical check you can repeat on a calculator.

limx → −1(2x³ − x + 4) Warm-up
Show answer and steps
2(−1)³ − (−1) + 4Polynomial, so plug in x = −1 directly= −2 + 1 + 4(−1)³ = −1= 3Answer
✓ Check: x = −1.001 gives 2.99499…; x = −0.999 gives 3.00499…
limx → 5x² − 25 over x − 5 Hole
Show answer and steps
0/0 at x = 5Plugging in gives 0/0, so do more work(x − 5)(x + 5) over x − 5Difference of squaresx + 5Cancel (x − 5); fine since x ≠ 55 + 5 = 10Now plug in
✓ Check: x = 4.999 gives 9.999; x = 5.001 gives 10.001.
limx → −4x² + 7x + 12 over x + 4 Hole
Show answer and steps
0/0 at x = −416 − 28 + 12 = 0 on top, 0 on the bottom(x + 3)(x + 4) over x + 4Factor: 3 · 4 = 12 and 3 + 4 = 7x + 3Cancel (x + 4)−4 + 3 = −1Plug in
✓ Check: x = −4.001 gives −1.001; x = −3.999 gives −0.999.
limx → 0(x + 3)² − 9 over x Expand first
Show answer and steps
0/0 at x = 09 − 9 = 0 on topx² + 6x + 9 − 9 over x = x(x + 6) over xExpand, then factor out xx + 6Cancel x0 + 6 = 6Plug in
✓ Check: x = 0.001 gives 6.001; x = −0.001 gives 5.999. (Spoiler: this is the slope of y = x² at x = 3, a preview of derivatives.)
limx → ∞4x − 7 over 2x + 9 At infinity
Show answer and steps
4 − 7/x over 2 + 9/xDivide top and bottom by x, the highest power4 − 0 over 2 + 07/x → 0 and 9/x → 0= 2Answer
✓ Check: x = 1,000,000 gives 1.9999875.
limx → ∞5x + 2 over x² + 1 At infinity
Show answer and steps
5/x + 2/x² over 1 + 1/x²Divide top and bottom by x²0 + 0 over 1 + 0Every term with x in the bottom → 0= 0Answer: the bottom grows much faster
✓ Check: x = 1,000 gives about 0.005002; x = 1,000,000 gives about 0.000005.
Let f(x) = x² for x < 2 and f(x) = 3x − 1 for x ≥ 2. Find limx → 2⁻f(x), limx → 2⁺f(x) and limx → 2f(x). One-sided
Show answer and steps
limx → 2⁻x² = 4Left side uses x²limx → 2⁺(3x − 1) = 5Right side uses 3x − 14 ≠ 5, so the limit does not existThe sides disagree
✓ Check: x = 1.999 gives 3.996001; x = 2.001 gives 5.003.
limx → 3⁻1 over x − 3 and limx → 3⁺1 over x − 3 Infinite
Show answer and steps
x slightly less than 3Then x − 3 is a tiny negative number1 ÷ (tiny negative) → −∞Left-hand limitx slightly more than 3Then x − 3 is a tiny positive number1 ÷ (tiny positive) → +∞Right-hand limit
✓ Check: x = 2.999 gives −1,000; x = 3.001 gives 1,000. The two-sided limit does not exist.
limx → 0√(x + 9) − 3 over x Challenge
Show answer and steps
0/0 at x = 0√9 − 3 = 0(√(x + 9) − 3)(√(x + 9) + 3) over x(√(x + 9) + 3)Multiply top and bottom by the conjugatex + 9 − 9 over x(√(x + 9) + 3) = 1 over √(x + 9) + 3(A − B)(A + B) = A² − B², then cancel x1 over 3 + 3 = 1 over 6Plug in x = 0
✓ Check: x = 0.001 gives 0.1666620…, and 1/6 = 0.1666666…

Section IXQuiz: eight quick questions

Choose one answer for each question, then press “Check my answers.” Your best score is saved only in this browser.

1. What is limx → 2(3x + 1)?

657The limit does not exist

2. What is limx → 1x² − 1x − 1?

201It does not exist, because 0/0 is undefined

3. A function is undefined at x = 3. What can you say about its limit as x → 3?

The limit must be 0The limit cannot exist, because there is no point at x = 3The limit equals f(3)It might exist; check values near 3

4. Near x = a, the left-hand limit is 4, the right-hand limit is 4, and f(a) = 7. What is the limit as x → a?

711It does not exist4

5. What is limx → ∞2x² + 35x² − x?

02/5∞2

6. With x in radians, what is limx → 0sin xx?

10It does not existπ/180

7. What happens to 1x as x → 0⁺ (from the right)?

It approaches 0It approaches 1It grows without bound: the limit is +∞It drops without bound: the limit is −∞

8. What is limx → −3x² + 5x + 6x + 3?

1−10−5

Scoring needs JavaScript turned on. Answer key: 1. 7; 2. 2; 3. It might exist; check values near 3; 4. 4; 5. 2/5; 6. 1; 7. It grows without bound: the limit is +∞; 8. −1.

Section XWhere limits came from

Limits are the foundation of calculus, but they came last. Newton and Leibniz built calculus in the late 1600s using “vanishingly small” quantities that were never quite defined. Newton’s Principia (1687) opens Book I with a section on what he called the method of “first and last ratios,” an early version of the limit idea.

Not everyone was convinced. In 1734 the philosopher George Berkeley published The Analyst, mocking those tiny quantities as “ghosts of departed quantities.” He had a point: calculus gave correct answers, but nobody could fully explain why.

The fix took about a century. In his Cours d’analyse of 1821, written for students at the École Polytechnique in Paris, Augustin-Louis Cauchy built calculus on limits as carefully as he could. Later in the 1800s, Karl Weierstrass and others tightened this into the precise “epsilon–delta” definition that college courses use today. The informal definition on this page is the same idea, minus the Greek letters.

NotesSources and credits

J. J. O’Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews: George Berkeley (biography), source of the Analyst date (1734).
George Berkeley, The Analyst (1734), section XXXV, edited by David R. Wilkins: online edition at Trinity College Dublin, source of the quotation (original capitals: “Ghosts of departed Quantities”).
MacTutor: Augustin-Louis Cauchy (biography), on the Cours d’analyse (1821), and Karl Weierstrass (biography), on his standards of rigor.
Judith V. Grabiner, “Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus,” American Mathematical Monthly 90, no. 3 (1983): 185–194, doi:10.2307/2975545.
Isaac Newton, The Mathematical Principles of Natural Philosophy, Motte (1729), 1846 American edition, Book I, Section I: “Of the method of first and last ratios of quantities” (Wikisource).

Credits: every graph and diagram on this page is original SVG, computed from the actual functions, made for Jugami. The interactive graphs are drawn live in your browser with plain JavaScript, no outside libraries.