A limit asks what the outputs approach as the inputs approach a point, never what happens at the point itself. A function can be undefined at 2, or defined as something absurd, and still have a limit of 4 there. Both sides must agree.
Step 1: Picture It
Start with something you already know.
A video with one frame missing
- A video shows a ball rolling slowly along a ruler.
- The frame at exactly 2 seconds got lost.
- Just before 2 seconds, the ball is a hair short of the 50 cm mark.
- Just after 2 seconds, it is a hair past the 50 cm mark.
- So you can tell where the ball was heading at 2 seconds, the 50 cm mark, without that frame.
A limit asks where the frames around a moment point; a lost frame, or a stray frame with a cat in it, changes nothing.
If the video cuts to a new scene at 2 seconds, the two sides point to different places, and there is no limit.
Where the picture falls short: a video has gaps between frames, but a function has an output at every x near the point.
The same idea with numbers. Take h(x) = 2x, and let x close in on 3 from both sides.
| x | h(x) |
|---|---|
| 2.9 | 5.8 |
| 2.99 | 5.98 |
| 3.01 | 6.02 |
| 3.1 | 6.2 |
From below and from above, the outputs close in on 6. That is the limit at 3, read from the neighbors without ever using x = 3 itself.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
f(x) = x + 3 for every x except 2, and f(2) = 9. Work out f(1.9), f(1.99), f(2.01) and f(2.1). What number are those four outputs closing in on?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Watch it approach
x = 1.9, 1.99, 1.999 give 3.9, 3.99, 3.999. x = 2.1, 2.01, 2.001 give 4.1, 4.01, 4.001. From both sides the values close in on 4.
The value there is irrelevant
Define f(2) = 100 and the limit at 2 is still 4. The limit is about the neighbors; the value is about the point. The exam tests exactly this distinction.
One-sided limits
x → 2⁻ approaches from the left, x → 2⁺ from the right. For f(x) = x + 1 when x < 2 and f(x) = 2x when x ≥ 2: left limit 3, right limit 4. They differ, so the limit does not exist.
Other failures
1/(x − 2) as x → 2: the values run away, +∞ from the right, −∞ from the left. No single number: the limit does not exist. sin(1/x) as x → 0 oscillates forever: no limit either.
Writing it
lim as x → 2 of (x² − 4)/(x − 2) = 4. The arrow is the whole idea: approaching, never arriving. Write DNE, does not exist, when there is no limit.
Why this is the foundation
The derivative is a limit of slopes, the integral a limit of sums. Both are 0/0 or ∞ × 0 at the point, and only the limit gets past that. Everything after this rests on it.
Step 3: Try It Yourself
Tap and try it out.
f(x) = x + 1 for x < 2, f(2) = 7, and f(x) = 2x for x > 2. Find the limit at 2, and at 1
- 1x → 2⁻: x + 1 → 3the left piece
Step 0 of 4
- Point(2, 0.50)
Step 4: In Real Life
A speedometer
A speedometer reports a limit: distance over an ever-shorter interval. No interval is zero long, yet the needle points at a number. That is where the ratio is heading.
Step 5: Watch an Example
One step at a time.
Watch Nadia Find the Limit of (x² − 36)/(x − 6) at 6
A fraction with a hole at 6, opened by factoring, then checked against its neighbors.
- Step 1
Substituting 6 gives (36 − 36)/(6 − 6), which is 0/0. That is not an answer. It is a signal: the top and the bottom both become zero at 6, so they share the factor (x − 6).
Watch Leo Test a Join, Then Finish It Yourself
f(x) = x² for x < 3, f(3) = 1, and f(x) = x + 6 for x > 3. Leo asks whether f has a limit at 3. He works the left side; the right side and the verdict are yours.
- Step 1
Like the join on the Try It Yourself page, the rule changes at 3, so Leo asks about each side on its own. Below 3 the function is x², above 3 it is x + 6, and the single value f(3) = 1 is set aside for now.
Watch Sam Meet a Number Over Zero
The limit of (x + 1)/(x − 3) as x approaches 3, where substitution gives something other than 0/0.
- Step 1
Substituting 3 gives (3 + 1)/(3 − 3), which is 4/0. This is not 0/0. The top is 4, not zero, so no shared factor is hiding and canceling cannot help.
Step 6: Your Turn
Practice makes it stick.
The Error Cell
Problem 1 of 14
A shipping sheet works out a charge in dollars with p(x) = (x² − 81)/(x − 9), where x is a parcel’s weight in pounds. The cell for exactly 9 pounds shows an error, because the formula gives 0 ÷ 0 there. What charge are the nearby cells heading toward at 9 pounds, in dollars?
dollars
The Two-Hour Line
Problem 2 of 14
A parking garage charges $8 for any stay under 2 hours, and $14 for a stay of 2 hours or more. Think of the charge as a function of the length of the stay. What is true about the charge at exactly 2 hours?
Where Is It Heading?
Substitute first, then decide whether you are done or need more work.
Problem 3 of 14
What is the limit of 2x + 5 as x approaches 4?
Problem 4 of 14
What is the limit of (x² − 64)/(x − 8) as x approaches 8?
Problem 5 of 14
f(x) = x + 4 for x < 5, and f(x) = 2x for x ≥ 5. What is the limit as x approaches 5 from the left?
Problem 6 of 14
g(x) = 3x − 1 for every x except 4, and g(4) = 40. What is the limit of g as x approaches 4?
Problem 7 of 14
What is the limit of (x² − 1)/(x + 1) as x approaches −1?
Whose Step?
One line on each page is a slip. Find it, then put it right.
Problem 8 of 14
Priya worked out the limit of (x² − 9)/(x − 3) as x approaches 3. One line is a slip. Which?
Priya’s working for the limit of (x² − 9)/(x − 3) as x → 3
- 1x = 3 gives (9 − 9)/(3 − 3) = 0/0
- 2x² − 9 = (x − 3)(x − 3)
- 3(x − 3)(x − 3)/(x − 3) = x − 3, for x ≠ 3
- Answerlimit = 3 − 3 = 0
Problem 9 of 14
Factored correctly, what is the limit of (x² − 9)/(x − 3) as x approaches 3?
Problem 10 of 14
Dev tested whether f has a limit at x = 1, where f(x) = x + 2 for x < 1 and f(x) = 4x for x ≥ 1. One line is a slip. Which?
Dev’s test for a limit at x = 1
- 1from the left: x + 2 → 1 + 2 = 3
- 2from the right: 4x → 4 × 1 = 4
- 3the sides differ, so the limit is their average: (3 + 4)/2
- Answerlimit = 3.5
Problem 11 of 14
Read correctly, what is true about the limit of f at x = 1?
Pick the Move
Choose the move. Neither of these needs working all the way through.
Problem 12 of 14
Substituting x = 3 into a fraction gives 0/0. What is the next move?
Problem 13 of 14
Lena wants to write that the limit of f as x approaches 3 is 9. For that to count, what must she show?
Same Limit?
Judge each one on its neighbors. Select every entry that fits.
Problem 14 of 14
Select every function whose limit as x approaches 2 is 4.
Step 7: Quick Check
Show what you know.
Question 1 of 1
Limit of (x² − 16)/(x − 4) as x approaches 4?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
f(x) = x² + k for x < 3, and f(x) = 2x + 5 for x ≥ 3. For which number k does f have a limit at x = 3?
Stretch 2 of 2
For exactly one number c, the limit of (x² − c)/(x − 5) as x approaches 5 is a number. Find c, then type that limit.
What You Learned
- A limit describes what the outputs approach, not what happens at the point.
- A limit exists only when both one-sided limits agree.
- 0/0 means more work is needed, not that the answer is zero.
- The derivative and the integral are both limits.
For the grown-up
The misconception this lesson targets is that a limit is just the function’s value said another way: that where the value is missing there is nothing to find, that where the value is present the limit must match it, and that 0/0 means the answer is 0. A student who holds it gets every plain substitution right and reads every hole and every jump as a misprint. It matters at once, because the derivative is a limit taken exactly where the quotient reads 0/0. The habit that fixes it is cheap: substitute first and read what comes back (a number, 0/0, or a number over 0), then take the neighbors on each side, and only then ask what the function does at the point, as a separate question.