Squeeze Theorem
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:
\[sin ( \frac{1}{x} ) \text{or} cos ( \frac{1}{x} )\]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 $x \rightarrow x_{0}$ exist and are equal to some real number $ℓ$: \(\underset{x \rightarrow x_{0}}{lim} g ( x ) = \underset{x \rightarrow x_{0}}{lim} h ( x ) = ℓ\)
Then, under these hypotheses, the function $g ( x )$ also admits a limit as $x \rightarrow x_{0}$, and that limit is: \(\underset{x \rightarrow x_{0}}{lim} f ( x ) = ℓ\)
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 $ℓ$, 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 $\epsilon > 0$ 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 $ℓ$ as $x \rightarrow x_{0}$. By assumption, we know that $\underset{x \rightarrow x_{0}}{lim} g ( x ) = ℓ$. 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: |
Similarly, since $\underset{x \rightarrow x_{0}}{lim} h ( x ) = ℓ$, there exists another positive number $\delta_{2}$ such that:
\[| h ( x ) - ℓ | < \epsilon \rightarrow ℓ - \epsilon < h ( x ) < ℓ + \epsilon\]| 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: |
Combining this with the bounds on $g ( x )$ and $h ( x )$, we obtain:
\[ℓ - \epsilon < f ( x ) < ℓ + \epsilon \rightarrow | f ( x ) - ℓ | < \epsilon\]Since this inequality holds for every $\epsilon > 0$, we conclude that:
\[\underset{x \rightarrow x_{0}}{lim} f ( x ) = ℓ\]Example
The subsequent example demonstrates how the theorem is applied to compute the following limit:
\[\underset{x \rightarrow 0}{lim} x \cdot sin ( \frac{1}{x} )\]The term $sin ( \frac{1}{x} )$ does not admit a limit as $x \rightarrow 0$, since it oscillates indefinitely between $- 1$ and $1$. However, for every real number $x \neq 0$, the following inequality holds:
\[- 1 \leq sin ( \frac{1}{x} ) \leq 1\]Multiplying the entire inequality by $x$, we obtain:
\[- | x | \leq x \cdot sin ( \frac{1}{x} ) \leq | x |\]| 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 $x \rightarrow 0$: |
Since $f ( x )$ is squeezed between two functions that both approach zero, we can apply the squeeze theorem and conclude that:
\[\underset{x \rightarrow 0}{lim} x \cdot sin ( \frac{1}{x} ) = 0\]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
- \(\text{1}. \underset{x \rightarrow + \infty}{lim} \frac{ln ( 3 + sin x )}{x^{3}}\) solution
- \(\text{2}. \underset{x \rightarrow 0}{lim} x^{4} \cdot cos ( \frac{2}{x} ) + 2\) solution
Exercise 1
Evaluate the following limit:
\[\underset{x \rightarrow + \infty}{lim} \frac{ln ( 3 + sin x )}{x^{3}}\]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:
\[2 \leq 3 + sin x \leq 4 \text{for all} x \in \mathbb{R}\]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:
\[\frac{log 2}{x^{3}} \leq \frac{ln ( 3 + sin x )}{x^{3}} \leq \frac{log 4}{x^{3}} \forall x > 0\]Since both bounding functions tend to zero as $x \rightarrow + \infty$, we apply the Squeeze Theorem and obtain:
\[\underset{x \rightarrow + \infty}{lim} \frac{ln ( 3 + sin x )}{x^{3}} = 0\]Exercise 2
Evaluate the following limit:
\[\underset{x \rightarrow 0}{lim} x^{4} \cdot cos ( \frac{2}{x} ) + 2\]To do so, we start by analyzing the behavior of the function $x^{4} \cdot cos ( \frac{2}{x} )$. We know that the cosine function is bounded between $- 1$ and $1$ for all real values:
\[- 1 \leq cos ( \frac{2}{x} ) \leq 1\]Multiplying all parts of this inequality by $x^{4}$, which is always non-negative, we get:
\[- x^{4} \leq x^{4} \cdot cos ( \frac{2}{x} ) \leq x^{4}\]Now we take the limit of the left and right bounds as $x \rightarrow 0$:
\[\underset{x \rightarrow 0}{lim} ( - x^{4} ) = 0 \underset{x \rightarrow 0}{lim} x^{4} = 0\]Therefore, by the Squeeze Theorem, we conclude:
\[\underset{x \rightarrow 0}{lim} x^{4} \cdot cos ( \frac{2}{x} ) = 0\]Now we return to the original expression:
\[\underset{x \rightarrow 0}{lim} ( x^{4} \cdot cos ( \frac{2}{x} ) + 2 )\]Since:
\[\underset{x \rightarrow 0}{lim} x^{4} \cdot cos ( \frac{2}{x} ) = 0\]we obtain
\[\underset{x \rightarrow 0}{lim} x^{4} \cdot cos ( \frac{2}{x} ) + 2 = 0 + 2 = 2\]Selected references
- MIT OpenCourseWare, C. Rodriguez. The Squeeze Theorem
- University of California, Berkeley, A. Vizeff. Limit Laws and the Squeeze Theorem