Squeeze Theorem for Limits

Introduction

In our last session, we developed an algebraic toolkit for solving limits that lead to the indeterminate form $\frac{0}{0}$ . By factoring and rationalizing, we could peel away the problematic terms and find the true value of the limit. However, not all functions are so easily tamed. What happens when a function oscillates wildly near a point, or is defined in a way that resists simple algebraic manipulation?

This is where we introduce a more subtle and elegant tool: the Squeeze Theorem, also known as the Sandwich Theorem or Pinching Theorem. This powerful principle allows us to find the limit of a complex function by trapping it between two simpler functions whose limits are known and equal.

In this lesson, you'll build an intuitive and then a formal understanding of the Squeeze Theorem. We will then apply it to solve several types of problems, including a cornerstone limit in calculus, $lim_{x \to 0} \frac{s i n ( x )}{x}$ , whose proof is a beautiful blend of geometry and analysis—exactly the kind of "out-of-the-box" reasoning valuable for the JEE.

The Intuition: Getting Squeezed into an Answer

Before we look at the formal mathematics, let's build a mental picture of what the Squeeze Theorem does. The name itself provides a great clue. Imagine a function $f (x)$ whose behavior is complicated, but you know it's always "sandwiched" between two other, simpler functions, say $g (x)$ and $h (x)$ .

A visual representation of the Sandwich (or Squeeze) Theorem. The function f(x) is always between h(x) and g(x). As x approaches 'a', both h(x) and g(x) converge to the same limit L, forcing f(x) to converge to L as well.

If we know that the two "slices of bread," $g (x)$ and $h (x)$ , are coming together at a single point, then the "filling," $f (x)$ , has no choice but to be squeezed toward that same point.

The following video from Khan Academy provides a wonderful, non-mathematical analogy to start with, before translating the idea into the language of functions and graphs.

Squeeze theorem or sandwich theorem | Limits | Differential Calculus | Khan Academy

This video builds the core intuition for the Squeeze Theorem, starting with a simple analogy and moving to a graphical representation.

Watch the first three segments to grasp the concept: The calorie analogy provides a common-sense foundation. Translating to functions shows how this looks on a graph. The limit condition explains the "squeeze" part, where the outer functions converge to the same value.

The Formal Statement and a First Application

Now that you have the intuition, let's look at the formal definition. The core idea remains the same, but it's expressed with the precision required for mathematical proofs.

Squeeze theorem and the limit of sin(x)/x

This article provides an excellent breakdown of the formal definition of the Squeeze Theorem.

Please read the first section, titled "The squeeze theorem". The author does a great job of parsing the formal statement and explaining what each part means. Then, read the subsection "Feeling the truth in your gut", which introduces a very useful slogan: Limits preserve non-strict inequalities.

In essence, the theorem gives us a three-step strategy:

Find two functions, $g (x)$ and $h (x)$ , that bound our target function $f (x)$ near the point of interest, $c$ . That is, $g (x) \leq f (x) \leq h (x)$ .
Show that the limits of the bounding functions are equal: $lim_{x \to c} g (x) = lim_{x \to c} h (x) = L$ .
Conclude that the limit of the target function must also be $L$ : $lim_{x \to c} f (x) = L$ .

Let's see this in action. A common source of functions that need squeezing involves trigonometric functions, because their values are naturally bounded. We know that for any angle $θ$ , $- 1 \leq sin (θ) \leq 1$ and $- 1 \leq cos (θ) \leq 1$ . This inequality is the starting point for many Squeeze Theorem problems.

Consider the limit $lim_{x \to 0} x^{2} cos (\frac{1}{x ^{2}})$ . Direct substitution fails because $cos (\infty)$ is undefined. However, we can squeeze it.

The following video works through this type of problem systematically.

Squeeze Theorem

This video by The Organic Chemistry Tutor provides a concise explanation and works through a classic example.

First, watch the brief introduction for a clear statement of the theorem: Main idea. Then, skip ahead and watch the worked example involving a cosine function: Example with cosine. Notice how the solution starts with the fundamental inequality -1 \le \cos(\dots) \le 1 and builds the target function from there.

The steps are:

Bound the core function: We know $- 1 \leq cos (\frac{1}{x ^{2}}) \leq 1$ .
Build the target function: Multiply the entire inequality by $x^{2}$ . Since $x^{2}$ is always non-negative, we don't need to worry about flipping the inequality signs. $- x^{2} \leq x^{2} cos (\frac{1}{x ^{2}}) \leq x^{2}$
Take the limits of the bounds:
- $lim_{x \to 0} (- x^{2}) = 0$
- $lim_{x \to 0} (x^{2}) = 0$
Apply the Squeeze Theorem: Since our function is trapped between two functions that both approach 0, its limit must also be 0.

The Geometric Proof of a Fundamental Limit: $lim_{x \to 0} \frac{s i n x}{x}$

Now for a truly classic application. The limit $lim_{x \to 0} \frac{s i n x}{x}$ is one of the most important in calculus. It's the foundation for finding the derivatives of all trigonometric functions. Direct substitution gives $\frac{0}{0}$ , but no simple algebraic trick like factoring works.

To solve this, we need a clever way to bound the function $\frac{s i n x}{x}$ . The method involves comparing the areas of three geometric shapes constructed within a unit circle. This is a beautiful piece of mathematical reasoning and a perfect example of the kind of creative problem-solving you'll encounter in advanced mathematics.

The reading below walks you through this elegant proof step-by-step. Take your time to follow the logic connecting the geometry to the algebra.

Squeeze theorem and the limit of sin(x)/x

This article provides a complete, step-by-step geometric proof of the limit of sin(x)/x.

Work through this step-by-step: First, understand why our simple approach fails. Read the section "Step 2", which shows that using -1 \le \sin(x) \le 1 directly leads to useless bounds. Next, see the "clever" geometric setup. Under the heading "Step 3: Get clever", study the diagram and the area formulas for the small triangle, the circular wedge (sector), and the large triangle. The core insight is that Area(small triangle) <= Area(wedge) <= Area(large triangle). Follow the algebraic manipulation. The rest of that section, from "Doing some algebra", shows how to transform the area inequality into bounds for \frac{\sin \theta}{\theta}. Pay close attention to how the author handles the cases for \theta > 0 and \theta < 0 to ensure the inequalities are correct. Finally, see the theorem in action. The concluding section, "Applying the squeeze theorem", takes the derived bounds, finds their limits, and formally concludes that \lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1.

Application to JEE-Style Problems

The Squeeze Theorem is also a go-to tool for problems involving the greatest integer function, denoted as $[x]$ , which gives the greatest integer less than or equal to $x$ . These problems are common in the JEE. The key is to use the defining property of the greatest integer function as our initial inequality:

x - 1 < [x] \leq x

Let's look at a typical problem structure.

Limits Using Sandwich Theorem & L'Hospital's Rule

This resource from askiitians demonstrates how to apply the Squeeze Theorem to a sum involving the greatest integer function, a common pattern in competitive exams.

Focus on Illustration 3. Read through the problem statement and the provided solution for the evaluation. Notice the strategy: An inequality is written for each term in the sum (e.g., rx-1 < [rx] \le rx). All these inequalities are summed up to create a bound for the entire numerator. The expression is divided by n^2, and the limits of the outer bounding functions are calculated as n \to \infty. Since both outer limits are equal, the Squeeze Theorem gives the final answer.

Conclusion

Today, we added a sophisticated and powerful technique to our calculus toolkit. Unlike direct algebraic assault, the Squeeze Theorem is about strategically trapping a difficult function to deduce its behavior.

Key Takeaways:

The Squeeze (or Sandwich) Theorem is used to find the limit of a function $f (x)$ by bounding it between two other functions, $g (x)$ and $h (x)$ , that converge to the same limit.
The strategy is to establish an inequality $g (x) \leq f (x) \leq h (x)$ and show that $lim_{x \to c} g (x) = lim_{x \to c} h (x) = L$ , which forces $lim_{x \to c} f (x) = L$ .
For oscillating functions like sine and cosine, the initial bound is typically $- 1 \leq sin (\dots) \leq 1$ or $- 1 \leq cos (\dots) \leq 1$ .
We used this theorem to derive the fundamental limit $lim_{x \to 0} \frac{s i n x}{x} = 1$ through a clever geometric argument involving areas.
For problems with the greatest integer function, the key inequality is $x - 1 < [x] \leq x$ .

The result $lim_{x \to 0} \frac{s i n x}{x} = 1$ is not just a one-off example; it's a foundational building block. In our next lesson, we will explore this and other standard limit forms involving trigonometric, exponential, and logarithmic functions. Mastering these will significantly speed up your ability to solve a wide range of limit problems you'll face in the JEE.

Recommended resources