Determining Limits Using the Squeeze Theorem
Why this matters
Imagine you're trying to figure out the exact location of your friend Jordan's drone as it lands. The drone's path is super erratic, zig-zagging back and forth. You can't track it directly. But, you have two friends, Priya and Marcus, who are walking steadily toward the same landing spot. Priya is always to the left of the drone's path, and Marcus is always to the right.
Even if you can't calculate the drone's wild path, you know that if Priya and Marcus both end up at the exact same spot on the ground, the drone must land there too. It has no other choice—it's been squeezed between them.
The Squeeze Theorem is the mathematical version of this. It's a powerful tool for finding limits of weird, wiggly functions that we can't solve any other way. We'll learn how to set up this "squeeze" and confidently find the limit.
Concept overview
flowchart TD
A[Start: Find lim f(x) as x->c] --> B{Can I use direct substitution or algebra?};
B -->|Yes| C[Solve directly. You're done!];
B -->|No| D{Is f(x) a product of a bounded function and a function going to 0?};
D -->|Yes| E[Identify the bounded part, e.g., sin(x) or cos(x)];
E --> F[Create an inequality, e.g., -1 <= sin(x) <= 1];
F --> G[Multiply/divide to make the middle f(x). Now you have g(x) <= f(x) <= h(x)];
G --> H{Do lim g(x) and lim h(x) equal the same value L?};
H -->|Yes| I[Conclusion: By Squeeze Theorem, lim f(x) = L];
H -->|No| J[Theorem doesn't apply. Re-check your g(x) and h(x)];
D -->|No| K[Try another limit technique];
Core explanation
Hello, future calculus masters! I'm Saavi, and I'm here to guide you through one of the most elegant tools in your limit toolkit: the Squeeze Theorem.
Sometimes, you'll hit a limit problem that resists all your usual methods. Direct substitution gives you an indeterminate form, and there's no obvious algebra to simplify it. These problems often involve a trigonometric function like sine or cosine, which oscillate back and forth, multiplied by another function.
Consider a limit like this:
lim (x→0) x² * sin(1/x)
If you try to plug in 0, you get 0 * sin(1/0), which is a mess. sin(1/0) is undefined and doesn't approach any single value. So, what do we do? We use the Squeeze Theorem.
The "Friend Sandwich" Analogy
Think of it like this: You're in the back of a car, squished between two friends, Carlos and Sofia. You are the middle function, f(x). Carlos, on your left, is the lower-bound function, g(x). Sofia, on your right, is the upper-bound function, h(x).
The car is heading to a big concert in downtown Dallas. The limit point, c, is the moment you all arrive at the venue's entrance. The limit value, L, is the specific parking spot you're all heading for.
If you know that Carlos and Sofia are both going to end up at Parking Spot B-4, and you are physically stuck between them, where are you going to end up? Parking Spot B-4. You have no choice.
That's the entire idea of the Squeeze Theorem.
The Three Conditions for the Squeeze Theorem
For the theorem to work, you have to prove three things. On the AP exam, you must show all three steps for full credit.
- 1The Squeeze ConditionYou must establish that your tricky function,
f(x), is truly stuck between your two chosen "friend" functions,g(x)andh(x), for all the x-values near your limit pointc.g(x) ≤ f(x) ≤ h(x) - 2The Agreement ConditionYou must show that both of your "friend" functions are heading to the exact same limit,
L, asxapproachesc.lim (x→c) g(x) = Landlim (x→c) h(x) = L - 3The ConclusionIf the first two conditions are met, you can confidently state that your function's limit must also be
L. Therefore,lim (x→c) f(x) = L.
Applying the Theorem: Step-by-Step
Let's go back to our problem: lim (x→0) x² * sin(1/x)
Step 1: Find your bounding functions.
This is where most students get stuck at first. You need to find the part of the function that you do know something about. Here, it's the sin(1/x) part.
No matter what you plug into a sine function—whether it's x, 1/x, or x²-4x+5—the output is always trapped between -1 and 1. This is the key!
So, we start with this known fact:
-1 ≤ sin(1/x) ≤ 1
Step 2: Build your inequality to match the original function.
Our original function isn't just sin(1/x), it's x² * sin(1/x). To get our inequality to match, we need to multiply all three parts by x².
Since x² is always positive (or zero), we don't have to worry about flipping the inequality signs.
(-1) * x² ≤ x² * sin(1/x) ≤ (1) * x²
-x² ≤ x² * sin(1/x) ≤ x²
Look at that! We just established our Squeeze Condition.
- Our
g(x)is-x². - Our
f(x)isx² * sin(1/x). - Our
h(x)isx².
Step 3: Check the limits of your bounding functions.
Now we check the Agreement Condition. Do our two simple, friendly functions g(x) and h(x) go to the same place as x approaches 0?
lim (x→0) -x² = -(0)² = 0
lim (x→0) x² = (0)² = 0
Yes! They both go to 0.
Step 4: State your conclusion.
We've shown that x² * sin(1/x) is squeezed between two functions that both approach 0. Therefore, by the Squeeze Theorem, our function must also approach 0.
Your formal answer on an exam would look like this:
Since
-x² ≤ x² * sin(1/x) ≤ x², andlim (x→0) -x² = 0andlim (x→0) x² = 0, then by the Squeeze Theorem,lim (x→0) x² * sin(1/x) = 0.
That's it. You've taken a complex, oscillating function and elegantly proven its limit without breaking a sweat. This method is your secret weapon for a very specific, but important, class of problems.
Worked examples
Let's walk through a couple of examples together. The key is to recognize the pattern: a bounded function (like sin or cos) multiplied by a function that goes to zero.
A Limit at Infinity
Problem: Find lim (x→∞) (cos(x)) / (x² + 5)
Step 1: Analyze the function and identify the "squeeze" opportunity.
As x goes to infinity, the denominator x² + 5 gets huge. The numerator, cos(x), just wiggles back and forth between -1 and 1. A number between -1 and 1 divided by an infinitely large number should be zero, but we need to prove it formally. This is a perfect setup for the Squeeze Theorem.
Step 2: Start with the bounded part of the function.
We know the range of the cosine function. No matter what x is, cos(x) is always between -1 and 1.
-1 ≤ cos(x) ≤ 1
Step 3: Build the inequality to match the original function.
Our function has cos(x) divided by x² + 5. So, let's divide all three parts of our inequality by x² + 5. Since x is going to infinity, x² + 5 is always a positive number, so we don't need to worry about flipping the inequality signs.
-1 / (x² + 5) ≤ cos(x) / (x² + 5) ≤ 1 / (x² + 5)
This is our Squeeze Condition. We have our g(x), f(x), and h(x).
Step 4: Find the limits of the outer functions. Now we check the Agreement Condition.
lim (x→∞) -1 / (x² + 5): Asxgets infinitely large, the denominatorx² + 5becomes infinitely large. -1 divided by a huge number is 0.lim (x→∞) 1 / (x² + 5): Similarly, 1 divided by a huge number is 0.
The limits match! Both g(x) and h(x) approach 0.
Step 5: State the conclusion.
Since our function is squeezed between two other functions that both approach 0 as x goes to infinity, our function's limit must also be 0.
Final Answer: Because -1/(x²+5) ≤ cos(x)/(x²+5) ≤ 1/(x²+5) and lim (x→∞) -1/(x²+5) = 0 and lim (x→∞) 1/(x²+5) = 0, by the Squeeze Theorem, lim (x→∞) cos(x)/(x²+5) = 0.
A Slightly Different Setup
Problem: Let f(x) be a function such that 4x - 9 ≤ f(x) ≤ x² - 4x + 7 for all x. Find lim (x→4) f(x).
Step 1: Analyze the problem.
This is a gift! The problem has already done the hard work of finding the bounding functions for you. You don't have to build the inequality; you just have to use it. Your g(x) is 4x - 9 and your h(x) is x² - 4x + 7.
This is where some students get confused. They think they need to find out what f(x) is. You don't! The whole point of the theorem is that we can find the limit of f(x) without ever knowing its exact formula.
Step 2: Find the limits of the given outer functions.
We just need to check the Agreement Condition. We'll find the limit of the left and right sides as x approaches 4.
-
lim (x→4) (4x - 9): We can use direct substitution here.4(4) - 9 = 16 - 9 = 7 -
lim (x→4) (x² - 4x + 7): Direct substitution works here, too.(4)² - 4(4) + 7 = 16 - 16 + 7 = 7
Step 3: State the conclusion.
The limits are the same! The lower bound and the upper bound both approach 7 as x approaches 4.
Final Answer: Since f(x) is bounded by 4x - 9 and x² - 4x + 7, and since lim (x→4) (4x - 9) = 7 and lim (x→4) (x² - 4x + 7) = 7, we can conclude by the Squeeze Theorem that lim (x→4) f(x) = 7.
Try it yourself
Ready to try on your own? Remember the process: find the bounded part, build the inequality, check the limits of the outer functions, and conclude.
Problem 1:
Find the limit: lim (x→0) x⁴ * cos(2/x)
Hint: What do you know about the range of cos(anything)? How can you use that to build an inequality? Remember that x⁴ is always non-negative.
Problem 2:
A function h(t) models the height of a kite in feet at time t in seconds. It is known that for t near 2 seconds, the height is bounded by two other functions: g(t) = -t² + 6t - 1 and k(t) = t² - 2t + 7. Find lim (t→2) h(t).
Hint: The problem has already given you the "squeeze." What's the only thing you need to check before you can draw your conclusion?
In simple terms, the Squeeze Theorem helps you find a tricky limit by "squeezing" it between two simpler functions that go to the same place.
- LIM-1.E: Determine the limits of functions using equivalent expressions for the function or the squeeze theorem.
- LIM-1.E.2
- The limit of a function may be found by using the squeeze theorem.
flowchart TD
A[Start: Find lim f(x) as x->c] --> B{Can I use direct substitution or algebra?};
B -->|Yes| C[Solve directly. You're done!];
B -->|No| D{Is f(x) a product of a bounded function and a function going to 0?};
D -->|Yes| E[Identify the bounded part, e.g., sin(x) or cos(x)];
E --> F[Create an inequality, e.g., -1 <= sin(x) <= 1];
F --> G[Multiply/divide to make the middle f(x). Now you have g(x) <= f(x) <= h(x)];
G --> H{Do lim g(x) and lim h(x) equal the same value L?};
H -->|Yes| I[Conclusion: By Squeeze Theorem, lim f(x) = L];
H -->|No| J[Theorem doesn't apply. Re-check your g(x) and h(x)];
D -->|No| K[Try another limit technique];
Read what Saavi narrates
Hi everyone, I'm Saavi, and today we're going to talk about one of my favorite tools in calculus, the Squeeze Theorem.
Imagine you're trying to track a friend's drone as it lands. Its path is super erratic, zig-zagging all over the place. But, you have two other friends, Priya and Marcus, walking steadily toward the same landing spot... Priya is always to the left of the drone, and Marcus is always to the right. Even if you can't calculate the drone's wild path, you know that if Priya and Marcus both end up at the exact same spot... the drone has to land there too. It's been squeezed between them.
That's the Squeeze Theorem. It's a strategy for finding the limit of a function that's hard to pin down. We trap our function between two simpler ones. If those two outer functions go to the same limit, the one in the middle has no choice but to go there too.
Let's work through a classic example. Find the limit as x approaches infinity of cosine of x, divided by x squared plus 5.
As x gets huge, the denominator... x squared plus 5... also gets huge. The numerator, cosine of x, just wiggles between negative 1 and 1. This is a perfect setup for the Squeeze Theorem.
We start with what we know about cosine. It's always between negative 1 and 1. So we can write the inequality... negative 1 is less than or equal to cosine of x, which is less than or equal to 1.
Now, we build up to our original function. We divide all three parts by x squared plus 5. That gives us... negative 1 over x squared plus 5... is less than or equal to... our original function... which is less than or equal to... 1 over x squared plus 5.
We've squeezed our function. Now for the final step: check the limits of the two outer functions as x goes to infinity. The limit of negative 1 over a huge number is zero. And the limit of 1 over a huge number is also zero.
They match! Since our function is squeezed between two functions that both go to zero, our limit must also be zero.
A common mistake here is picking the wrong bounds. Some students write that sine or cosine is between 0 and 1. But that's wrong... they can be negative! Always start with negative 1 to 1.
The Squeeze Theorem feels a little strange at first, but it's an incredibly powerful and logical tool. Once you get the hang of setting up the inequality, you'll find these problems become some of the most satisfying to solve. You've got this.
The entire theorem depends on the two "friends" going to the *same* spot. If `lim g(x) = 5` and `lim h(x) = 8`, you can't conclude anything about `f(x)`.
Always calculate both limits separately and show they are equal before stating your conclusion.
`sin(x)` and `cos(x)` can be negative. Their range is `[-1, 1]`. Using the wrong bounds will lead to an incorrect inequality.
Always start your inequality for sine or cosine as `-1 ≤ ... ≤ 1`.
If you multiply `-2 < 5` by `-1`, you get `2 > -5`. The inequality flips. The same happens with variables. If `x` could be negative, multiplying by `x` requires careful case analysis (which is why we often multiply by `x²` or `|x|`).
When building your inequality, if you multiply by a term that could be negative (like `x` as `x→0`), it's safer to multiply by a term you know is positive, like `x²`. For the problems you'll see, this is almost always the case.
On an AP Free Response Question (FRQ), you get points for showing your reasoning. You must explicitly state the inequality and show that the limits of the outer functions are equal.
Write out the full three-part justification: 1) the inequality, 2) the limits of the outer functions, and 3) the conclusion.
It's like using a sledgehammer to crack a nut. If you can find a limit by direct substitution or factoring, do it. It's faster and less prone to error.
Always try direct substitution first. Reserve the Squeeze Theorem for its specific use case: weird, oscillating functions that you can bound.