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Mastering Calculus: The Squeeze Theorem Explained (With Examples)

Calculus provides plenty of algebraic tricks for evaluating limits—factoring, multiplying by conjugates, and dividing by the highest power. But eventually, you will encounter a function that completely defies algebra. It might oscillate infinitely or contain a messy trigonometric expression that cannot be simplified.

When algebraic manipulation fails, you need a logical workaround. This is where the Squeeze Theorem (often called the Sandwich Theorem or Pinching Theorem) comes to the rescue. Instead of evaluating the difficult function directly, you trap it between two simple functions and force it to yield its limit.

Here is exactly how the Squeeze Theorem works, why it is a brilliant mathematical tool, and several step-by-step examples of how to apply it.

What is the Squeeze Theorem?

The theorem states that if a tricky function is always trapped between two well-behaved functions, and those two outer functions converge to the exact same limit at a specific point, the trapped function must also converge to that same limit.

Mathematically, suppose that for all x near a (except possibly at a itself):
g(x) \le f(x) \le h(x)
If we know that the limits of the outer functions are equal:
\lim_{x \to a} g(x) = L, \lim_{x \to a} h(x) = L
Then, by the Squeeze Theorem, the limit of the inner function must also be L:
\lim_{x \to a} f(x) = L
Think of it like two pieces of bread ($g(x)$ and h(x)) squeezing a piece of meat ($f(x)$). If both pieces of bread are moving toward the exact same location, the meat has nowhere else to go—it must end up in that location, too.

5 Examples of the Squeeze Theorem in Action

The key to using this theorem is building an inequality. You usually start with a known bounded function (like sine or cosine) and algebraically build the rest of your function around it.

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Example 1: The Classic Oscillating Function

If you try to evaluate this limit directly, you get 0 \cdot \text{undefined}, which isn’t helpful:
\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)
We know that the sine of any angle is always trapped between -1 and 1:
-1 \le \sin\left(\frac{1}{x}\right) \le 1
Now, multiply all three parts of the inequality by x^2 (which is always positive, so the inequality signs don’t flip) to match our original function:
-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2
Take the limit of the two outer functions as x approaches 0:
\lim_{x \to 0} (-x^2) = 0, \lim_{x \to 0} (x^2) = 0
Since both the “bottom bread” and “top bread” go to 0, the inner function is squeezed to 0:
\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0

Example 2: Limits at Infinity

The Squeeze Theorem is incredibly useful for evaluating trigonometric limits at infinity, which otherwise oscillate endlessly without settling on a single value.
\lim_{x \to \infty} \frac{\cos(x)}{x}
Start with the known boundary of the cosine function:
-1 \le \cos(x) \le 1
Divide the entire inequality by x (assuming x > 0 as it approaches positive infinity):
-\frac{1}{x} \le \frac{\cos(x)}{x} \le \frac{1}{x}
Take the limit of the outer pieces as x \to \infty:
\lim_{x \to \infty} \left(-\frac{1}{x}\right) = 0$\lim_{x \to \infty} \left(\frac{1}{x}\right) = 0
Because both outer limits are 0, the middle limit is forced to be 0:
\lim_{x \to \infty} \frac{\cos(x)}{x} = 0

Example 3: Working with Abstract Inequalities

Sometimes, a problem will hand you the inequality directly, testing your understanding of the concept rather than your algebra skills.

Suppose you are given the following inequality for all x:
4x - 9 \le f(x) \le x^2 - 4x + 7
Find \lim_{x \to 4} f(x).

You don’t know what f(x) is, but you don’t need to. Just evaluate the limit of the left side and the right side as x approaches 4.

Left side:

\lim_{x \to 4} (4x - 9) = 16 - 9 = 7
Right side:

\lim_{x \to 4} (x^2 - 4x + 7) = 16 - 16 + 7 = 7
Since both outer limits equal 7, the limit of the trapped function must be 7:
\lim_{x \to 4} f(x) = 7

Example 4: Dealing with Exponents

You can use the Squeeze Theorem with squared trigonometric functions. Because any real number squared is positive, the lower bound changes.
\lim_{x \to \infty} \frac{\sin^2(x)}{x^2 + 1}
Because -1 \le \sin(x) \le 1, squaring it means it will fluctuate between 0 and 1:
0 \le \sin^2(x) \le 1
Divide the inequality by x^2 + 1:
0 \le \frac{\sin^2(x)}{x^2 + 1} \le \frac{1}{x^2 + 1}
Take the limits of the endpoints as x approaches infinity. The left side is a constant 0. For the right side:
\lim_{x \to \infty} \frac{1}{x^2 + 1} = 0
Both sides go to 0, so the limit of the middle function is 0.

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Example 5: Absolute Value Limits

You can also bound a function using absolute values to deal with signs cleanly.
\lim_{x \to 0} \vert{}x\vert{} \cos\left(\frac{\pi}{x}\right)
Start with the bounds of cosine:
-1 \le \cos\left(\frac{\pi}{x}\right) \le 1
Multiply everything by \vert{}x\vert{} (which is positive, keeping the inequality intact):
-\vert{}x\vert{} \le \vert{}x\vert{} \cos\left(\frac{\pi}{x}\right) \le \vert{}x\vert{}
Evaluate the limits of the two outer functions as x approaches 0:
\lim_{x \to 0} -\vert{}x\vert{} = 0, \lim_{x \to 0} \vert{}x\vert{} = 0
Both outer limits converge to 0, meaning the inner limit is squeezed to 0.

Why This Rule Matters

While it might feel like a clever trick to solve a few highly specific math homework problems, the Squeeze Theorem is a major pillar of theoretical calculus.

It is the primary tool used by mathematicians to prove one of the most important fundamental limits in all of calculus: \lim_{x \to 0} \frac{\sin(x)}{x} = 1. Without the Squeeze Theorem, we wouldn’t be able to prove that specific limit, and without that limit, we wouldn’t be able to find the derivatives of trigonometric functions like sine and cosine. By mastering how to bound and squeeze unpredictable functions, you are interacting with the very logic that makes advanced calculus work.

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