Always substitute first. For a continuous function the limit is the value. 0 ÷ 0 reports that a shared factor is hiding the answer: factor and cancel, rationalise with a conjugate, or clear a compound fraction. The Squeeze Theorem handles what algebra cannot.
Step 1: Picture It
Start with something you already know.
Walking between two friends
- Two friends walk you down a hallway, one on each side.
- You may wobble and weave, but only in the space between them.
- Both friends are walking to the same door.
- As they near it, the space between them shrinks.
- So you reach that door too, however much you wobbled.
Near the end of this lesson, the Squeeze Theorem says the same about a function trapped between two others.
If both bounds head for the same number, the trapped function must head there too.
It only works if the friends reach the same door, and you stay between them near it.
x² sin(1/x) wobbles, but it always stays between −x² and x². Its room to wobble is the gap, 2x².
| x | Room to wobble, 2x² |
|---|---|
| 0.1 | 0.02 |
| 0.01 | 0.0002 |
| 0.001 | 0.000002 |
The room shrinks to 0, so x² sin(1/x) is forced to 0.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
What is the limit of (x² − 4)/(x² − 2x) as x approaches 2?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Factor and cancel
(x² − x − 6)/(x² − 4) as x → 2: (x − 2)(x − 3)/((x − 2)(x + 2)) = (x − 3)/(x + 2) → −1/4. Canceling changes the function only at the one point the limit ignores.
A compound fraction
(1/x − 1/3)/(x − 3) as x → 3: the top is (3 − x)/(3x), so the whole is −(x − 3)/(3x(x − 3)) = −1/(3x) → −1/9.
The Squeeze Theorem
If g is trapped between f and h near a point, and f and h share a limit there, g is forced to that same limit. −x² ≤ x² sin(1/x) ≤ x², and both bounds → 0, so x² sin(1/x) → 0 as x → 0, though it oscillates wildly.
The special limits
sin x/x → 1 and (1 − cos x)/x → 0 as x → 0. Check: sin 0.01/0.01 = 0.99998. They appear in every derivation of the trigonometric derivatives and on the exam every year.
Using them
sin 3x/x as x → 0: write it as 3 · sin 3x/(3x). As x → 0, 3x → 0, so sin 3x/(3x) → 1 and the limit is 3. Match the inside of the sine to the denominator.
Not all zeros are equal
5/0 is not indeterminate: the values run away and the limit does not exist. ∞/∞ and 0 × ∞ are indeterminate; 5/0 and 5/∞ are not. Name the form before choosing a method.
Step 3: Try It Yourself
Tap and try it out.
Find the limit of (x² − x − 6)/(x² − 4) as x → 2, and of sin 3x/x as x → 0
- 1x = 2: (4 − 2 − 6)/(4 − 4) = 0/0indeterminate
Step 0 of 4
- Point(1, 2)
Step 4: In Real Life
A formula at its edge
Average cost at zero items is 0/0. Factor and cancel to see the cost of the first unit. The Squeeze Theorem does the same job for an oscillating signal a sensor cannot resolve.
Step 5: Watch an Example
One step at a time.
Watch Maya Clear a Root with a Conjugate
Maya needs the limit of (√(x + 5) − 3)/(x − 4) as x approaches 4. A root sits on the top, so there is no bracket to factor.
- Step 1
Substituting 4 gives (√9 − 3)/(4 − 4) = (3 − 3)/0, which is 0/0. That is a signal, not an answer: something shared by the top and the bottom is hiding the limit.
Watch Leo Start a Conjugate, Then Finish It Yourself
Leo needs the limit of (√x − 2)/(x − 4) as x approaches 4. He reads the form and picks the tool; the algebra and the answer are yours.
- Step 1
Substituting 4 gives (√4 − 2)/(4 − 4) = (2 − 2)/0, which is 0/0. A root sits on the top, so Leo multiplies the top and the bottom by its conjugate, √x + 2.
Watch Ana Match a Special Limit
Ana needs the limit of sin 5x/(2x) as x approaches 0. The inside of the sine and the bottom do not match yet.
- Step 1
Substituting 0 gives sin 0/(2 × 0) = 0/0, so the limit needs more than arithmetic.
Step 6: Your Turn
Practice makes it stick.
The Filling Tank
Problem 1 of 14
A tank fills so that after t minutes it holds t² + 6t liters. Between minute 2 and minute t, the average filling rate is (t² + 6t − 16)/(t − 2) liters per minute. What does that average head toward as t approaches 2?
liters per minute
Splitting the Pizza
Problem 2 of 14
Friends split a $60 pizza bill evenly, so with x friends each pays 60/x dollars. Starting from 4 friends, each share drops by (15 − 60/x)/(x − 4) dollars per extra friend. The formula works for any x, not just whole numbers. What does it head toward as x approaches 4?
dollars per friend
Open the Limit
Substitute first. When it gives 0/0, pick the tool the shape calls for.
Problem 3 of 14
What is the limit of (x² + 2x − 8)/(x² − 4) as x approaches 2? Type it as a decimal.
Problem 4 of 14
What is the limit of (x − 9)/(√x − 3) as x approaches 9?
Problem 5 of 14
What is the limit of (1/x − 1/2)/(x − 2) as x approaches 2? Type it as a decimal.
Problem 6 of 14
What is the limit of sin 4x/(2x) as x approaches 0?
Problem 7 of 14
For every x near 0, 5 − x² ≤ g(x) ≤ 5 + x². What is the limit of g as x approaches 0?
Whose Step?
One line on each page is a slip. Find it, then put it right.
Problem 8 of 14
Ravi found the limit of (x² − 16)/(x² − 3x − 4) as x approaches 4. One line is a slip. Which?
Ravi’s working for the limit of (x² − 16)/(x² − 3x − 4) as x → 4
- 1x = 4: (16 − 16)/(16 − 12 − 4) = 0/0
- 2(x − 4)(x + 4)/((x − 4)(x − 1))
- 3(x + 4)/(x − 1), for x ≠ 4
- Answerlimit = 8/3
Problem 9 of 14
Factored correctly, what is the limit of (x² − 16)/(x² − 3x − 4) as x approaches 4? Type it as a decimal.
Problem 10 of 14
Sam found the limit of sin 6x/(2x) as x approaches 0. One line is a slip. Which?
Sam’s working for the limit of sin 6x/(2x) as x → 0
- 1x = 0: sin 0/0 = 0/0
- 2sin 6x/(2x) = 6 · sin 6x/(6x)
- 3sin 6x/(6x) → 1 as x → 0
- Answerlimit = 6 × 1 = 6
Problem 11 of 14
Rewritten so it still equals sin 6x/(2x), what is the limit as x approaches 0?
Pick the Move
Name the form, then choose. Neither of these needs working all the way through.
Problem 12 of 14
Substituting 0 into (√(x + 16) − 4)/x gives 0/0. Which move opens it up?
Problem 13 of 14
Substituting 2 into (x + 3)/(x − 2) gives 5/0. What should be written next?
Same Limit?
Substitute first, then use the tool each shape calls for. Select every entry that fits.
Problem 14 of 14
Select every limit below that must equal 2.
Step 7: Quick Check
Show what you know.
Question 1 of 2
Limit of (x² − 1) ÷ (x − 1) as x approaches 1?
Question 2 of 2
What does the Squeeze Theorem require?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
For exactly one number c, (√(x + c) − 5)/x reads 0/0 at x = 0. Find c, then type the limit as x approaches 0, as a decimal.
Stretch 2 of 2
For which number k does sin(kx)/(3x) approach 4 as x approaches 0?
What You Learned
- Substitute first; a number is the answer.
- 0 ÷ 0 means factor, rationalise, or clear the fractions.
- The Squeeze Theorem forces a limit when a function is trapped between two others.
- sin x/x → 1 and (1 − cos x)/x → 0; match the inside to the denominator to use them.
For the grown-up
The misconception this lesson targets is that 0/0 is an answer: that it means the limit is 0, or 1, or that there is no limit, and its twin, that any zero on the bottom calls for the same algebra. 0/0 says only that the top and the bottom share something that vanishes at the point; the work is to find it and remove it, by factoring, by a conjugate, or by clearing a compound fraction. A non-zero number over 0 is a different report entirely: the values run away and the limit does not exist, and no factoring changes that. The slip that travels with it is a rewrite that quietly changes the value, such as a bracket that does not multiply back to the original, or a special limit used before the bottom matches the inside of the sine. The habit that prevents all three is cheap: substitute, name the form, choose the tool, and multiply every rewrite back out before trusting it.