Squeeze Theorem

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What is the Squeeze Theorem

The Squeeze Theorem, also referred to as the Sandwich Theorem, provides a method for determining the limit of a function when direct evaluation is challenging or when the function displays complex oscillatory behaviour near a specific point. This theorem is frequently applied to functions involving sine and cosine, particularly when these trigonometric terms exhibit oscillatory behaviour that precludes straightforward limit evaluation, such as:

\sin\left(\frac{1}{x}\right) \quad \text{or} \quad \cos\left(\frac{1}{x}\right)

In these situations, the function is constrained between two other functions with known and equal limits, which facilitates the evaluation of the target limit.

Statement

Let $x_0 \in \mathbb{R} \cup { \pm\infty }$ be a limit point, meaning that every neighborhood of $x_0$ contains at least one point of the domain different from $x_0$. Let $f$, $g$, and $h$ be real-valued functions defined on a neighborhood $I$ of $x_0$. Assume that for every $x \in I$, the following inequality holds:

g(x) \leq f(x) \leq h(x)

Also assume that the limits of $f(x)$ and $h(x)$ as $x \to x_0$ exist and are equal to some real number $\ell$:

\lim_{x \to x_0} g(x) = \lim_{x \to x_0} h(x) = \ell

Then, under these hypotheses, the function $g(x)$ also admits a limit as $x \to x_0$, and that limit is:

\lim_{x \to x_0} f(x) = \ell

In the graph, the black curve representing $f(x)$ lies entirely between the lower bound $g(x)$ and the upper bound $h(x)$. As both bounding functions tend to $\ell$, the function $f(x)$ is forced to approach the same limit.

This demonstrates the geometric intuition underlying the theorem: if a function is bounded above and below by two functions that both converge to the same value, then it must converge to that value.

Proof of the Squeeze Theorem

Let $\varepsilon > 0$ be arbitrary. Our objective is to prove that the function $f(x)$, which is bounded between $g(x)$ and $h(x)$, tends to the same limit $\ell$ as $x \to x_0$. By assumption, we know that $\lim_{x \to x_0} g(x) = \ell$. This means that there exists a positive number $\delta_1$ such that for every $x$ sufficiently close to $x_0$ (specifically, for all $x$ with $0 < |x - x_0| < \delta_1$), we have:

|g(x) - \ell| < \varepsilon \quad \to \quad \ell - \varepsilon < g(x) < \ell + \varepsilon

Similarly, since $\lim_{x \to x_0} h(x) = \ell$, there exists another positive number $\delta_2$ such that:

|h(x) - \ell| < \varepsilon \quad \to \quad \ell - \varepsilon < h(x) < \ell + \varepsilon

Now let $\delta = \min(\delta_1, \delta_2)$. Then for every $x$ such that $0 < |x - x_0| < \delta$, both inequalities above are satisfied. But $f(x)$ is squeezed between $g(x)$ and $h(x)$, so:

g(x) \leq f(x) \leq h(x)

Combining this with the bounds on $g(x)$ and $h(x)$, we obtain:

\ell - \varepsilon < f(x) < \ell + \varepsilon \quad \to \quad |f(x) - \ell| < \varepsilon

Since this inequality holds for every $\varepsilon > 0$, we conclude that:

\lim_{x \to x_0} f(x) = \ell

Example

The subsequent example demonstrates how the theorem is applied to compute the following limit:

\lim_{x \to 0} x \cdot \sin\left( \frac{1}{x} \right)

The term $\sin\left( \frac{1}{x} \right)$ does not admit a limit as $x \to 0$, since it oscillates indefinitely between $-1$ and $1$. However, for every real number $x \neq 0$, the following inequality holds:

-1 \leq \sin\left( \frac{1}{x} \right) \leq 1

Multiplying the entire inequality by $x$, we obtain:

-|x| \leq x \cdot \sin\left( \frac{1}{x} \right) \leq |x|

Indeed, when $x > 0$, the inequality is preserved, while for $x < 0$, the inequality is reversed, but the absolute value ensures that the comparison remains symmetric with respect to zero. Now observe that both bounding functions $-|x|$ and $|x|$ tend to zero as $x \to 0$:

\lim_{x \to 0} -|x| = 0 \qquad \lim_{x \to 0} |x| = 0

Since $f(x)$ is squeezed between two functions that both approach zero, we can apply the squeeze theorem and conclude that:

\lim_{x \to 0} x \cdot \sin\left( \frac{1}{x} \right) = 0

In many instances, when an oscillating function is multiplied by a power of $x$ that approaches zero, the overall limit is zero. This result arises because the oscillation remains bounded, as demonstrated by sine and cosine functions, which are always confined between $-1$ and $1$. Conversely, the factor $x^n$ approaches zero rapidly enough to dominate the oscillation, causing the entire product to converge to zero.

Exercises: compute the following limits using the Squeeze Theorem

-

\text{1. } \quad \lim_{x \to +\infty} \frac{\ln(3 + \sin x)}{x^3}

solution

-

\text{2. } \quad \lim_{x \to 0} x^4 \cdot \cos\left(\frac{2}{x}\right) + 2

solution

Exercise 1

Evaluate the following limit:

\lim_{x \to +\infty} \frac{\ln(3 + \sin x)}{x^3}

To begin, observe that the sine function is always bounded between $-1$ and $1$ for all real $x$, so we can write:

-1 \leq \sin x \leq 1

From the previous inequality, we can write:

2 \leq 3 + \sin x \leq 4 \quad \text{for all } x \in \mathbb{R}

Now, since the logarithmic function is strictly increasing, we have:

\log 2 \leq \log(3 + \sin x) \leq \log 4

We now divide all parts of the inequality by $x^3$ obtaining:

\frac{\log 2}{x^3} \leq \frac{\ln(3 + \sin x)}{x^3} \leq \frac{\log 4}{x^3} \quad \forall \, x > 0

Since both bounding functions tend to zero as $x \to +\infty$, we apply the Squeeze Theorem and obtain:

\lim_{x \to +\infty} \frac{\ln(3 + \sin x)}{x^3} = 0

Exercise 2

Evaluate the following limit:

\lim_{x \to 0} x^4 \cdot \cos\left( \frac{2}{x} \right) + 2

To do so, we start by analyzing the behavior of the function $x^4 \cdot \cos\left( \frac{2}{x} \right)$. We know that the cosine function is bounded between $-1$ and $1$ for all real values:

-1 \leq \cos\left( \frac{2}{x} \right) \leq 1

Multiplying all parts of this inequality by $x^4$, which is always non-negative, we get:

-x^4 \leq x^4 \cdot \cos\left( \frac{2}{x} \right) \leq x^4

Now we take the limit of the left and right bounds as $x \to 0$:

\lim_{x \to 0} (-x^4) = 0 \qquad \lim_{x \to 0} x^4 = 0

Therefore, by the Squeeze Theorem, we conclude:

\lim_{x \to 0} x^4 \cdot \cos\left( \frac{2}{x} \right) = 0

Now we return to the original expression:

\lim_{x \to 0} \left( x^4 \cdot \cos\left( \frac{2}{x} \right) + 2 \right)

Since:

\lim_{x \to 0} x^4 \cdot \cos\left( \frac{2}{x} \right) = 0

we obtain

\lim_{x \to 0} x^4 \cdot \cos\left( \frac{2}{x} \right) + 2 = 0 + 2 = 2

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