Calculus presents us with various useful computational tricks, especially in the field of limits. When faced with oscillating functions or functions with undefined points, taking the limit can become a difficult task. Luckily, The Squeeze, or Sandwich, Theorem is just the trick for dealing with tricky functions such as these.
The Squeeze Theorem Definition
The Squeeze Theorem is a limit evaluation method where we "squeeze" an indeterminate limit between two simpler ones. The "squeezed" or "bounded" function approaches the same limit as the other two functions surrounding it.
More precisely, the Squeeze Theorem states that for functions \(f\), \(g\), and \(h\) such that:
\[g(x) \leq f(x) \leq h(x)\]
if
\[lim_{x \rightarrow A } g(x)= lim_{x \rightarrow A } h(x)=L\]
for a constant \(L\), then:
\[lim_{x \rightarrow A} f(x)=L\]
Fig. 1. As \(f(x)\) is "squeezed" between \(g(x)\) and \(h(x)\), the Squeeze Theorem can be applied to evaluate the limit of \(f(x)\) at \(x = 0\).
The Squeeze Theorem Proof
Informal Proof
Simply put, \(f(x)\) is "squeezed" between \(g(x)\) and \(h(x)\). As \(g(x)\) and \(h(x)\) are equal at the point A such that \(g(A)=h(A)=L\), then \(f(A)=L\) , as there is no room between the other two functions for \(f\) to take on any other value.
Formal Proof
We will assume that
- \(g(x)\leq f(x) \leq h(x)\)everywhere in the domain of the functions
- \(lim_{x \rightarrow A} g(x)=lim_{x \rightarrow A} h(x)=L\)
Upon these assumptions, you want to prove that:
\[lim_{x \rightarrow A} f(x)=L\]
See the image below for a visual explanation of the variables!
Let an arbitrary epsilon such that \(\epsilon > 0\) be known. To prove the Squeeze Theorem, we must find a delta \(\delta > 0\) such that \(|f(x)-L|< \epsilon\) whenever \(0< |x-A|< \delta\) where L is the evaluation of the limit as \(x\) approaches the point \(A\).
Now \(\lim_{x \rightarrow A} g(x)=L\) by definition, so there must exist some \(\delta_g > 0\) such that:
\(|g(x)-L|< \epsilon\) for all \(0 < |x-A|< \delta_g\):
Using absolute value laws
\[-\epsilon + L < g(x) < \epsilon + L\]
for all
\[0<|x-A|<\delta_g\]
Similarly, \(lim_{x \rightarrow A} h(x)=L\) by definition, so there must exist some \(\delta_g > 0\)such that
\(|h(x)-L|< \epsilon\) for all \(0 < |x-A|< \delta_g\).
Using absolute value laws
\[- \epsilon + L < h(x)< \epsilon + L \]
for all
\[0<|x-A|<\delta_g\]
Fig. 2. Visual explanation of the geometric derivation of (1) and (2).
Since \(g(x)\leq f(x) \leq h(x)\) for all \(x\) on some open interval containing \(A\), there must exist some \(\delta_f > 0\) such that
(3) \(g(x) \leq f(x) \leq h(x)\) for all \(0< |x-A|< \delta_f\)
Where \(\delta = min (\delta_g, \delta_h, \delta_f)\), then by (1), (2), and (3)
\[-\epsilon + L < g(x) \leq f(x) \leq h(x) < L + \epsilon\] for all \(0<|x-A|<\delta\)
Thus
\(-\epsilon< f(x)-L < \epsilon \) for all/ \(0<|x-A|<\delta\)
Using absolute value laws
\( |f(x)-L| < \epsilon\) for all \(0<|x-A|<\delta\)
Then, by definition:
\[lim_{x \rightarrow A} f(x)=L\]
When to use the Squeeze Theorem Formula
The Squeeze Theorem should be used as a last resort. When solving limits, one should always try to solve through algebraic or simple manipulation first. If algebra fails, the Squeeze Theorem may be a viable option for limit solving.
Indeed, to calculate \(lim_{x \rightarrow A} f(x)\), we must first find two functions \(g(x)\) and \(h(x)\) that bound \(f(x)\) and such that:
\[lim_{x \rightarrow A}g(x)=lim_{x \rightarrow A} h(x)\]
The Squeeze Theorem cannot be applied if the limits of the bounding functions are not equal.
Examples of Evaluating Limits using the Squeeze Theorem
Let's start with a simple example!
Use the Squeeze Theorem to evaluate
\[lim_{x \rightarrow 0} x^2 \cos \left( \dfrac{1}{x^2} \right)\]
When we plug in \(x = 0\), we are met with an undefined form \(cos \left( \frac{1}{0} \right)\). This is a perfect candidate for the Squeeze Theorem!
This is an example of a general scenario: the Squeeze Theorem can be applied to find the limit of trigonometric functions damped by polynomial terms.
The general strategy for solving these kinds of Squeeze Theorem problems is to- start with the trigonometric function,\(f(x)=\cos \left( \dfrac{1}{x^2} \right)\) in this case- build up to the function in the problem; here it is \(f(x)=x^2 \cos \left( \dfrac{1}{x^2} \right)\)Let's see how this is done!
- Step 1: Make a double-sided inequality to bound the trigonometric function based on the nature of the cosine function.
- We know that the cosine function oscillates on the closed interval \([-1, 1]\), i.e. \(-1 \leq cos(x) \leq 1\)
- In graphing \(\cos \left( \frac{1}{x^2} \right)\), we find that f also oscillates on the closed interval \([-1, 1]\), i.e. \(-1 \leq \cos \left( \frac{1}{x^2} \right)\leq 1\)
It is essential to know that the range of \(\cos\) (anything) and \(\sin\) (anything) will always be \([-1, 1]\) (as long as it is not translated up/down or vertically stretched/compressed)!
Fig. 3. Example 1 graph.
- Step 2: Modify the inequality as needed to bound the function of the problem:
- Our function is \(f(x)=x^2 \cos \left( \frac{1}{x^2} \right)\) so we multiply our double-sided inequality by \(x^2\) and get:
\[-x^2\leq x^2 \cos \left( \frac{1}{x^2} \right)\leq x^2\]
- Step 3: Verify that the bounding functions have the same limit.
- Now that our function is bounded, we must verify that \(lim_{x \rightarrow 0}-x^2= lim_{x \rightarrow 0} x^2\) in order to apply the Squeeze Theorem: \(lim_{x \rightarrow 0}-x^2= lim_{x \rightarrow 0} x^2=0\)
- Step 4: Apply the Squeeze Theorem
- Since \(lim_{x \rightarrow 0}-x^2= lim_{x \rightarrow 0} x^2=0\), then by the Squeeze Theorem\[lim_{x \rightarrow 0} x^2 \cos \left( \frac{1}{x^2}\right)=0\]
Now, let's try something a bit more complex.
Find \(lim_{x \rightarrow - \infty} \dfrac{7x^2-\sin(5x)}{x^2+15}\)
When we plug in \(- \infty\), we are left with the indeterminate form \(\dfrac{\infty}{\infty}\). Again,
since a trigonometric function appears, this is a perfect candidate for the Squeeze Theorem!Following the same strategy as before, start with the trigonometric function \(f(x)=\sin(5x)\), and build up to
\[f(x)=\dfrac{7x^2-\sin(5x)}{x^2+15}\]
Let's take a look!
- Step 1: Make a double-sided inequality to bound the trigonometric function based on the nature of the sine function.
- We know that sine behaves like cosine in that it oscillates on the closed interval \([-1, 1]\).
- Looking at the image below, when we graph \(-\sin(5x)\), we find that:
\[-1 \leq -\sin(5x) \leq 1\]
Fig. 4. Example 2 graph.
- Step 2: Modify the inequality as needed to bound the function of the problem
- Our function is:\[f(x)=lim_{x \rightarrow - \infty} \dfrac{7x^2-\sin(5x)}{x^2+15}\] and our double-sided inequality is: \[-1 \leq -\sin(5x) \leq 1\]
- Add \(7x^2\) to get \[7x^2-1 \leq 7x^2-\sin(5x) \leq 7x^2+1\]
- Multiply by \(\frac{1}{x^2+15}\) to get \[\dfrac{7x^2-1}{x^2+15} \leq \dfrac{7x^2-\sin(5x)}{x^2+15} \leq \dfrac{7x^2+1}{x^2+15} \]
- Step 3: Now that our function is bounded, we must verify that: \[lim_{x \rightarrow - \infty }\dfrac{7x^2-1}{x^2+15}=lim_{x \rightarrow -\infty } \dfrac{7x^2+1}{x^2+15}\] This in order to apply the Squeeze Theorem.
- Again, when we try to plug in \(- \infty\), we are met with an indeterminate form. However, this time we can use algebraic manipulation to solve.
- Multiply both limits by: \[\dfrac{\frac{1}{x^2}}{\frac{1}{x^2}}\] to get \[lim_{x \rightarrow -\infty \dfrac{7- \frac{1}{x^2}}{1+ \frac{15}{x^2}} }\] and \[lim_{x \rightarrow -\infty \dfrac{7+ \frac{1}{x^2}}{1+ \frac{15}{x^2}} }\]
- Now, when we plug in \(-\infty\), we get \(\dfrac{7-0}{1+0}\) and \(\dfrac{7+0}{1+0}\) respectively leaving: \[lim_{x \rightarrow -\infty} \dfrac{7x^2-1}{x^2-15}=lim_{x \rightarrow -\infty} \dfrac{7x^2+1}{x^2-15}=7\]
- Step 4: Apply the Squeeze Theorem
- Since \[lim_{x \rightarrow -\infty} \dfrac{7x^2-1}{x^2-15}=lim_{x \rightarrow -\infty} \dfrac{7x^2+1}{x^2-15}=7\], then by the Squeeze Theorem: \[lim_{x \rightarrow -\infty } \dfrac{7x^2-\sin(5x)}{x^2+15}=7\]
The Squeeze Theorem - Key takeaways
- The Squeeze Theorem is a last resort method for solving limits that cannot be solved through algebraic manipulation.
- The Squeeze Theorem states that for functions \(f\), \(g\), and \(h\) such that: \[g(x) \leq f(x) \leq h(x) \], if: \[lim_{x \rightarrow A} g(x)= lim_{x \rightarrow A} h(x)=L\]
for a constant \(L\), then: \[lim_{x \rightarrow A} f(x)=L\]
- If \(lim_{x \rightarrow A} g(x) \neq lim_{x \rightarrow A} h(x)\), the Squeeze Theorem cannot be applied
- The general strategy for solving problems containing trigonometric functions is to start with the trig function, then build up to the function in the question!
- A step-by-step procedure for the Squeeze Theorem is:
- Step 1: Make a double-sided inequality based on the nature of \(f(x)\).
- Step 2: Modify the inequality as needed
- Step 3: Solve the limits on both sides of the inequality, ensuring that they are equal
- Step 4: Apply the Squeeze Theorem - the limit of \(f(x)\) is equal to the bounding limits
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