Calculus I: The Squeeze Theorem and Proofs

In engineering, you often encounter functions whose behavior is complex or oscillatory near a point, making their limits difficult to evaluate directly. The Squeeze Theorem (or Sandwich Theorem) is an indispensable tool for taming these unruly functions. It allows you to find the limit of a complicated expression by "squeezing" it between two simpler, better-understood functions whose limits you already know. Mastering this theorem is crucial for analyzing waveforms, signals, and systems where direct substitution fails.

Understanding the Squeeze Theorem

The Squeeze Theorem provides a formal method for finding the limit of a function $f(x)$ by comparing it to two other functions, $g(x)$ and $h(x)$. The theorem states: If $g(x)\le f(x)\le h(x)$ for all $x$ in an interval around a point $c$ (except possibly at $c$ itself), and if ${\lim }_{x\to c}g(x)={\lim }_{x\to c}h(x)=L$, then it must also be true that ${\lim }_{x\to c}f(x)=L$.

Think of it as a car traveling between two police cruisers on a highway. If both cruisers converge to the same exit at 65 mph, the car trapped between them has no choice but to follow and take the same exit at the same speed. The power of the theorem lies in its logic: you don't need to know the exact path of the middle function, only that it is forever bounded by two functions sharing a common limit. This is especially useful when $f(x)$ involves troublesome components like $\sin (1/x)$ or $\cos (1/x)$ that oscillate wildly as $x$ approaches zero.

The Art of Identifying Bounding Functions

The core skill in applying the Squeeze Theorem is constructing effective bounding functions, $g(x)$ and $h(x)$. Your goal is to find simpler functions that definitively trap your target function $f(x)$. A universal starting point is to use known inequalities, with the most important being $-1\le \sin (\theta )\le 1$ and $-1\le \cos (\theta )\le 1$ for any real angle $\theta$.

Consider the classic engine for oscillation: $x^2\sin (1/x)$. Direct substitution as $x\to 0$ is impossible because $\sin (1/x)$ oscillates infinitely. However, we know $\sin (1/x)$ is bounded. We can build an inequality: $$-1\le \sin (1/x)\le 1.$$ Multiplying all parts of this inequality by $x^2$ (which is non-negative) gives: $$-x^2\le x^2\sin (1/x)\le x^2.$$ We have now successfully squeezed our complex function, $f(x)=x^2\sin (1/x)$, between $g(x)=-x^2$ and $h(x)=x^2$. Since both $g(x)$ and $h(x)$ clearly approach $0$ as $x\to 0$, the Squeeze Theorem forces the middle function to do the same: ${\lim }_{x\to 0}{x}^2\sin (1/x)=0$. The bounding functions you choose don't need to be the "tightest" possible bounds; they just need to be simple and share the same, easily computable limit.

A Foundational Proof: ${\lim }_{x\to 0}\frac{\sin x}{x}=1$

This limit is the cornerstone of trigonometric calculus and its proof is the quintessential application of the Squeeze Theorem. You cannot use direct substitution (0/0) or simple algebraic manipulation. The proof relies on a geometric inequality constructed using a unit circle.

For a small angle $x$ (in radians) where $0<x<\pi /2$, consider the following areas on a unit circle:

1. Area of triangle $OAP$ = $\frac{1}{2}\cdot 1\cdot \sin x=\frac{\sin x}{2}$.
2. Area of circular sector $OAP$ = $\frac{1}{2}\cdot 1^2\cdot x=\frac{x}{2}$.
3. Area of triangle $OAQ$ = $\frac{1}{2}\cdot 1\cdot \tan x=\frac{\tan x}{2}$.

Visually, you can see that: Area of $△OAP$ $\le$ Area of sector $OAP$ $\le$ Area of $△OAQ$. This translates to the inequality: $$\frac{\sin x}{2}\le \frac{x}{2}\le \frac{\tan x}{2}.$$ Multiplying through by 2 gives: $\sin x\le x\le \tan x$. For the squeeze, we need to isolate $\frac{\sin x}{x}$. Taking the reciprocal of $x\le \tan x=\frac{\sin x}{\cos x}$ and carefully manipulating the inequalities (considering the signs for negative $x$ as well) leads to the classic double inequality: $$\cos x\le \frac{\sin x}{x}\le 1.$$ Now, we have our squeeze. The function $f(x)=\frac{\sin x}{x}$ is bounded below by $g(x)=\cos x$ and above by $h(x)=1$. Critically, ${\lim }_{x\to 0}\cos x=1$ and ${\lim }_{x\to 0}1=1$. Therefore, by the Squeeze Theorem, ${\lim }_{x\to 0}\frac{\sin x}{x}=1$.

Constructing a Formal Squeeze Theorem Proof

To write a rigorous proof, follow this structured, four-step process. Let's apply it to prove ${\lim }_{x\to 0}x\cos (1/x)=0$.

Step 1: Identify the Target and the Point of Approach. We want to prove ${\lim }_{x\to 0}f(x)=0$, where $f(x)=x\cos (1/x)$. The problematic component is $\cos (1/x)$.

Step 2: Establish the Bounding Inequality. Use the known bound for cosine: $-1\le \cos (1/x)\le 1$. To incorporate the $x$ factor, multiply the entire inequality by $x$. However, you must consider the sign of $x$. For $x>0$, multiplication preserves inequality: $$x\cdot (-1)\le x\cdot \cos (1/x)\le x\cdot 1\Rightarrow -x\le x\cos (1/x)\le x.$$ For $x<0$, multiplying by a negative number reverses the inequalities: $$x\cdot (-1)\ge x\cdot \cos (1/x)\ge x\cdot 1\Rightarrow -x\ge x\cos (1/x)\ge x.$$ Notice that in both cases, $f(x)$ is between $x$ and $-x$. We can write a single, clean inequality that holds for all $x\ne 0$: $$-|x|\le x\cos (1/x)\le |x|.$$ Here, $g(x)=-|x|$ and $h(x)=|x|$.

Step 3: Verify the Limits of the Bounding Functions. Evaluate: ${\lim }_{x\to 0}-|x|=0$ and ${\lim }_{x\to 0}|x|=0$. Both limits equal $L=0$.

Step 4: Apply the Squeeze Theorem. Since $g(x)\le f(x)\le h(x)$ for all $x\ne 0$ and ${\lim }_{x\to 0}g(x)={\lim }_{x\to 0}h(x)=0$, the Squeeze Theorem concludes that ${\lim }_{x\to 0}f(x)=0$. Your proof is complete.

Combining with Limit Laws and Tackling Oscillations

The Squeeze Theorem rarely exists in isolation. You must often use it in tandem with other limit laws to solve more complex problems. A common pattern is to use the theorem to find the limit of a problematic part of a function, then use the limit laws (like the product or sum law) to find the limit of the whole expression.

Consider evaluating ${\lim }_{x\to 0}\left[3+x^2\sin \left(\frac{1}{x}\right)\right]$. You cannot directly apply the limit sum law because the limit of the second term doesn't seem to exist. However, use the Squeeze Theorem first on the oscillatory part. As shown earlier, $-x^2\le x^2\sin (1/x)\le x^2$, so ${\lim }_{x\to 0}{x}^2\sin (1/x)=0$. Now, you can legally apply the sum law: $$\underset{x\to 0}{\lim }\left[3+x^2\sin \left(\frac{1}{x}\right)\right]=\underset{x\to 0}{\lim }3+\underset{x\to 0}{\lim }{x}^2\sin \left(\frac{1}{x}\right)=3+0=3.$$ This strategy is powerful for oscillating functions near a point. Functions like $\sin (1/x)$ or $\cos (1/x)$ have no limit as $x\to 0$ on their own. But when multiplied by a function that approaches zero (like $x$, $x^2$, or $\sqrt{x}$), the entire product can be squeezed to zero. The bounding functions absorb the oscillation, allowing you to pin down the limit.

Common Pitfalls

1. Ignoring the Sign When Multiplying Inequalities: This is the most frequent error. When you multiply an inequality by a quantity, you must check if it's positive (inequality stays the same) or negative (inequality reverses). The safest way to avoid this is to use absolute values to create a single, sign-independent inequality, as shown with $|x|$.
2. Assuming the Bounding Functions Share a Limit: The theorem's conclusion is only valid if $\lim g(x)=\lim h(x)$. If your chosen $g(x)$ and $h(x)$ have different limits, the theorem tells you nothing about $f(x)$. Always compute these two limits explicitly as the first step in your verification.
3. Applying the Theorem Backwards: You cannot conclude that the bounding functions have a certain limit because the middle function does. The logic flows only one way: from the bounds to the squeezed function.
4. Overlooking the "Interval Around c" Condition: The inequality $g(x)\le f(x)\le h(x)$ must hold for all $x$ in a deleted interval surrounding point $c$. It's not enough for it to be true at a few points. For the standard applications with $x\to 0$, using global bounds like $-1\le \sin (\theta )\le 1$ satisfies this condition automatically.

Summary

• The Squeeze Theorem allows you to find the limit of a function $f(x)$ by trapping it between two functions $g(x)$ and $h(x)$ that converge to the same value $L$ at a point $c$.
• The key technique is to use known inequalities (especially $-1\le \sin \theta ,\cos \theta \le 1$) to construct simple bounding functions for complex, oscillatory terms.