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Practice Problems

Use the squeeze theorem to find the following limits (if possible).

\(\textbf{1)}\) \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x}\)

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The answer is \(0\)

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Step 1: \(-1 \le \sin{x} \le 1\)

Step 2: \(\displaystyle\frac{-1}{x} \le \displaystyle\frac{\sin{x}}{x} \le \frac{1}{x}\)

Step 3: \(\displaystyle\lim_{x\to \infty}\frac{-1}{x} \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x} \le \displaystyle\lim_{x\to \infty}\frac{1}{x}\)

Step 4: \(\displaystyle\frac{-1}{\infty} \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x} \le \frac{1}{\infty}\)

Step 5: \(0 \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x} \le 0\)

Step 6: \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x}=0\) by Squeeze Theorem

\(\textbf{2)}\) \(\displaystyle\lim_{x\to 0} \sin{\frac{1}{x}}+1\)

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The answer is Does Not Exist (DNE)

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Step 1: \(-1 \le \sin{x} \le 1\)

Step 2: \(-1\le \sin{\frac{1}{x}} \le 1\)

Step 3: \(-1+1\le \sin{\frac{1}{x}}+1 \le 1+1\)

Step 4: \(0\le \sin{\frac{1}{x}}+1 \le 2\)

Step 5: \(\displaystyle\lim_{x\to 0}0\le \displaystyle\lim_{x\to 0}\sin{\frac{1}{x}}+1 \le \displaystyle\lim_{x\to 0}2\)

Step 6: \(0\le \displaystyle\lim_{x\to 0}\sin{\frac{1}{x}}+1 \le 2\)
Step 7: \(\displaystyle\lim_{x\to 0}\sin{\frac{1}{x}}+1\) does not exist since \(0 \neq 2\)

\(\textbf{3)}\) \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x-1}\)

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The answer is \(0\)

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Step 1: \(-1 \le \sin{x} \le 1\)

Step 2: \(\displaystyle\frac{-1}{x-1} \le \displaystyle\frac{\sin{x}}{x-1} \le \frac{1}{x-1}\)

Step 3: \(\displaystyle\lim_{x\to \infty}\frac{-1}{x-1} \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x-1} \le \displaystyle\lim_{x\to \infty}\frac{1}{x-1}\)

Step 4: \(\displaystyle\frac{-1}{\infty+1} \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x+1} \le \frac{1}{\infty+1}\)

Step 5: \(0 \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x+1} \le 0\)

Step 6: \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x+1}=0\) by Squeeze Theorem

\(\textbf{4)}\) \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{3}\)

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The answer is Does Not Exist (DNE)

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Step 1: \(-1 \le \sin{x} \le 1\)

Step 2: \(\displaystyle\frac{-1}{3} \le \displaystyle\frac{\sin{x}}{3} \le \frac{1}{3}\)

Step 3: \(\displaystyle\lim_{x\to \infty}\frac{-1}{3} \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{3} \le \displaystyle\lim_{x\to \infty}\frac{1}{3}\)

Step 4: \(\displaystyle\frac{-1}{3} \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{3} \le \frac{1}{3}\)

Step 5: \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x}\) does not exist since \(\displaystyle\frac{-1}{3} \neq \displaystyle\frac{1}{3}\)

\(\textbf{5)}\) \(\displaystyle\lim_{x\to -\infty}\frac{\cos{x}}{x^2}\)

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The answer is \(0\)

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Step 1: \(-1 \le \cos{x} \le 1\)

Step 2: \(\displaystyle\frac{-1}{x^2} \le \displaystyle\frac{\cos{x}}{x^2} \le \frac{1}{x^2}\)

Step 3: \(\displaystyle\lim_{x\to \infty}\frac{-1}{x^2} \le \displaystyle\lim_{x\to \infty}\frac{\cos{x}}{x^2} \le \displaystyle\lim_{x\to \infty}\frac{1}{x^2}\)

Step 4: \(\displaystyle\frac{-1}{\infty^2} \le \displaystyle\lim_{x\to \infty}\frac{\cos{x}}{x^2} \le \frac{1}{\infty^2}\)

Step 5: \(0 \le \displaystyle\lim_{x\to \infty}\frac{\cos{x}}{x^2} \le 0\)

Step 6: \(\displaystyle\lim_{x\to \infty}\frac{\cos{x}}{x^2}=0\) by Squeeze Theorem

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Continuity on Graphs}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Piecewise Functions- Limits and Continuity}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Infinite Limits}\)
\(\,\,\,\,\,\,\,\,\displaystyle \lim_{x\to 4^{+}} \frac{5}{x-4}\)

\(\bullet\text{ Limits at Infinity}\)
\(\,\,\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5x^2+2x-10}{3x^2+4x-5}\)

\(\bullet\text{ Trig Limits}\)
\(\,\,\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin \theta}{\theta}=1\)

In Summary

The Squeeze theorem, also known as the Sandwich theorem or the Pinching theorem, is a mathematical concept that allows us to figure out the value of a function if we can “sandwich” it between 2 other functions. Essentially, the Squeeze theorem states that if two functions “sandwich” a third function, then the value of the third function must lie between the values of the other two functions.

In formal terms, the Squeeze theorem states that if \(g(x) \le f(x) \le h(x)\) for all \(x\) in a certain interval, and if both \(g(x)\) and \(h(x)\) approach a limit L as \(x\) approaches a, then \(f(x)\) must also approach the same limit L as \(x\) approaches a. This theorem is extremely useful for determining the value of limits that are difficult or impossible to compute directly.

The Squeeze theorem is typically covered in a multivariable calculus course, usually at the undergraduate level. It is an important tool for understanding the behavior of functions and their limits, and it is often used in conjunction with other limit theorems.

In addition to its use in multivariable Calculus, the Squeeze theorem has applications in other areas of mathematics, such as real analysis, complex analysis, and differential equations. It is a fundamental tool for understanding the behavior of functions and their limits, and it is an essential concept for anyone interested in mathematical analysis.

Topics that use Squeeze theorem

Limits of sequences: The squeeze theorem can be used to prove the existence of limits of sequences. For example, if you have a sequence \(b_n\) such that \(a_n \lt b_n \lt c_n\) for all n, and you can show that the limits of the sequences \(a_n\) and \(c_n\) both exist and are equal to L, then you can conclude that the limit of the sequence \(b_n\) also exists and is equal to L.

Taylor series: The squeeze theorem can be used to prove the convergence of Taylor series. For example, if you have a function f(x) that can be approximated by a Taylor series expansion, and you can find two functions \(g(x)\) and \(h(x)\) such that \(g(x) /le f(x) /le h(x)\) for all x in the interval of convergence, and the Taylor series expansions of \(g(x)[]/latex and [latex]h(x)\) both converge, then you can conclude that the Taylor series expansion of \(f(x)\) also converges.

Improper integrals: The squeeze theorem can be used to prove the convergence of improper integrals.

Power series: The squeeze theorem can be used to prove the convergence of power series.

Fourier series: The squeeze theorem can be used to prove the convergence of Fourier series.

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