The Squeeze Theorem is used to find limits when a function is trapped between two other functions that approach the same value. It is especially useful for limits involving bounded trig functions like \(\sin{x}\) and \(\cos{x}\) multiplied or divided by expressions that approach \(0\). These problems include limits at infinity, limits as \(x\to0\), and cases where the Squeeze Theorem does not prove a limit.
Practice Problems
Use the squeeze theorem to find the following limits (if possible).
\(\textbf{1)}\) \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x}\)
The answer is \(0\)
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: \(0 \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x} \le 0\)
Step 5: \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x}=0\) by Squeeze Theorem
\(\textbf{2)}\) \(\displaystyle\lim_{x\to 0} \sin{\frac{1}{x}}+1\)
The answer is Does Not Exist (DNE)
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: \(\displaystyle\lim_{x\to 0}0\le \displaystyle\lim_{x\to 0}\left(\sin{\frac{1}{x}}+1\right) \le \displaystyle\lim_{x\to 0}2\)
Step 6: \(0\le \displaystyle\lim_{x\to 0}\left(\sin{\frac{1}{x}}+1\right) \le 2\)
Step 7: The Squeeze Theorem does not prove a limit here since \(0 \neq 2\). In fact, \(\displaystyle\lim_{x\to 0}\left(\sin{\frac{1}{x}}+1\right)\) does not exist.
\(\textbf{3)}\) \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x-1}\)
The answer is \(0\)
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: \(0 \le \displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x-1} \le 0\)
Step 5: \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{x-1}=0\) by Squeeze Theorem
\(\textbf{4)}\) \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{3}\)
The answer is Does Not Exist (DNE)
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: The Squeeze Theorem does not prove a limit here since \(\displaystyle\frac{-1}{3} \neq \displaystyle\frac{1}{3}\). In fact, \(\displaystyle\lim_{x\to \infty}\frac{\sin{x}}{3}\) does not exist.
\(\textbf{5)}\) \(\displaystyle\lim_{x\to -\infty}\frac{\cos{x}}{x^2}\)
The answer is \(0\)
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: \(0 \le \displaystyle\lim_{x\to -\infty}\frac{\cos{x}}{x^2} \le 0\)
Step 5: \(\displaystyle\lim_{x\to -\infty}\frac{\cos{x}}{x^2}=0\) by Squeeze Theorem
\(\textbf{6)}\) \(\displaystyle\lim_{x\to 0}x^2\sin{\left(\frac{1}{x}\right)}\)
The answer is \(0\)
Step 1: \(-1 \le \sin{\left(\frac{1}{x}\right)} \le 1\)
Step 2: \(-x^2 \le x^2\sin{\left(\frac{1}{x}\right)} \le x^2\)
Step 3: \(\displaystyle\lim_{x\to 0}(-x^2) \le \displaystyle\lim_{x\to 0}x^2\sin{\left(\frac{1}{x}\right)} \le \displaystyle\lim_{x\to 0}x^2\)
Step 4: \(0 \le \displaystyle\lim_{x\to 0}x^2\sin{\left(\frac{1}{x}\right)} \le 0\)
Step 5: \(\displaystyle\lim_{x\to 0}x^2\sin{\left(\frac{1}{x}\right)}=0\) by Squeeze Theorem
\(\textbf{7)}\) \(\displaystyle\lim_{x\to 0}x\cos{\left(\frac{1}{x}\right)}\)
The answer is \(0\)
Step 1: \(-1 \le \cos{\left(\frac{1}{x}\right)} \le 1\)
Step 2: \(-|x| \le x\cos{\left(\frac{1}{x}\right)} \le |x|\)
Step 3: \(\displaystyle\lim_{x\to 0}(-|x|) \le \displaystyle\lim_{x\to 0}x\cos{\left(\frac{1}{x}\right)} \le \displaystyle\lim_{x\to 0}|x|\)
Step 4: \(0 \le \displaystyle\lim_{x\to 0}x\cos{\left(\frac{1}{x}\right)} \le 0\)
Step 5: \(\displaystyle\lim_{x\to 0}x\cos{\left(\frac{1}{x}\right)}=0\) by Squeeze Theorem
\(\textbf{8)}\) \(\displaystyle\lim_{x\to \infty}\frac{\cos{x}}{2x+1}\)
The answer is \(0\)
Step 1: \(-1 \le \cos{x} \le 1\)
Step 2: \(\displaystyle\frac{-1}{2x+1} \le \displaystyle\frac{\cos{x}}{2x+1} \le \frac{1}{2x+1}\)
Step 3: \(\displaystyle\lim_{x\to \infty}\frac{-1}{2x+1} \le \displaystyle\lim_{x\to \infty}\frac{\cos{x}}{2x+1} \le \displaystyle\lim_{x\to \infty}\frac{1}{2x+1}\)
Step 4: \(0 \le \displaystyle\lim_{x\to \infty}\frac{\cos{x}}{2x+1} \le 0\)
Step 5: \(\displaystyle\lim_{x\to \infty}\frac{\cos{x}}{2x+1}=0\) by Squeeze Theorem
\(\textbf{9)}\) \(\displaystyle\lim_{x\to 0}x^3\sin{\left(\frac{2}{x^2}\right)}\)
The answer is \(0\)
Step 1: \(-1 \le \sin{\left(\frac{2}{x^2}\right)} \le 1\)
Step 2: \(-|x^3| \le x^3\sin{\left(\frac{2}{x^2}\right)} \le |x^3|\)
Step 3: \(\displaystyle\lim_{x\to 0}(-|x^3|) \le \displaystyle\lim_{x\to 0}x^3\sin{\left(\frac{2}{x^2}\right)} \le \displaystyle\lim_{x\to 0}|x^3|\)
Step 4: \(0 \le \displaystyle\lim_{x\to 0}x^3\sin{\left(\frac{2}{x^2}\right)} \le 0\)
Step 5: \(\displaystyle\lim_{x\to 0}x^3\sin{\left(\frac{2}{x^2}\right)}=0\) by Squeeze Theorem
\(\textbf{10)}\) Suppose \(-4x^2 \le f(x) \le 7x^2\). Find \(\displaystyle\lim_{x\to 0}f(x)\).
The answer is \(0\)
Step 1: \(-4x^2 \le f(x) \le 7x^2\)
Step 2: \(\displaystyle\lim_{x\to 0}(-4x^2) \le \displaystyle\lim_{x\to 0}f(x) \le \displaystyle\lim_{x\to 0}7x^2\)
Step 3: \(0 \le \displaystyle\lim_{x\to 0}f(x) \le 0\)
Step 4: \(\displaystyle\lim_{x\to 0}f(x)=0\) by Squeeze Theorem
\(\textbf{11)}\) \(\displaystyle\lim_{x\to \infty}\frac{2\sin{x}}{x^2}\)
The answer is \(0\)
Step 1: \(-1 \le \sin{x} \le 1\)
Step 2: \(-2 \le 2\sin{x} \le 2\)
Step 3: \(\displaystyle\frac{-2}{x^2} \le \displaystyle\frac{2\sin{x}}{x^2} \le \frac{2}{x^2}\)
Step 4: \(\displaystyle\lim_{x\to \infty}\frac{-2}{x^2} \le \displaystyle\lim_{x\to \infty}\frac{2\sin{x}}{x^2} \le \displaystyle\lim_{x\to \infty}\frac{2}{x^2}\)
Step 5: \(0 \le \displaystyle\lim_{x\to \infty}\frac{2\sin{x}}{x^2} \le 0\)
Step 6: \(\displaystyle\lim_{x\to \infty}\frac{2\sin{x}}{x^2}=0\) by Squeeze Theorem
\(\textbf{12)}\) \(\displaystyle\lim_{x\to 0}x^4\cos{\left(\frac{5}{x}\right)}\)
The answer is \(0\)
Step 1: \(-1 \le \cos{\left(\frac{5}{x}\right)} \le 1\)
Step 2: \(-x^4 \le x^4\cos{\left(\frac{5}{x}\right)} \le x^4\)
Step 3: \(\displaystyle\lim_{x\to 0}(-x^4) \le \displaystyle\lim_{x\to 0}x^4\cos{\left(\frac{5}{x}\right)} \le \displaystyle\lim_{x\to 0}x^4\)
Step 4: \(0 \le \displaystyle\lim_{x\to 0}x^4\cos{\left(\frac{5}{x}\right)} \le 0\)
Step 5: \(\displaystyle\lim_{x\to 0}x^4\cos{\left(\frac{5}{x}\right)}=0\) by Squeeze Theorem
\(\textbf{13)}\) \(\displaystyle\lim_{x\to \infty}\frac{5\cos{x}}{\sqrt{x}}\)
The answer is \(0\)
Step 1: \(-1 \le \cos{x} \le 1\)
Step 2: \(-5 \le 5\cos{x} \le 5\)
Step 3: \(\displaystyle\frac{-5}{\sqrt{x}} \le \displaystyle\frac{5\cos{x}}{\sqrt{x}} \le \frac{5}{\sqrt{x}}\)
Step 4: \(\displaystyle\lim_{x\to \infty}\frac{-5}{\sqrt{x}} \le \displaystyle\lim_{x\to \infty}\frac{5\cos{x}}{\sqrt{x}} \le \displaystyle\lim_{x\to \infty}\frac{5}{\sqrt{x}}\)
Step 5: \(0 \le \displaystyle\lim_{x\to \infty}\frac{5\cos{x}}{\sqrt{x}} \le 0\)
Step 6: \(\displaystyle\lim_{x\to \infty}\frac{5\cos{x}}{\sqrt{x}}=0\) by Squeeze Theorem
\(\textbf{14)}\) \(\displaystyle\lim_{x\to 0}\left|x\right|\sin{\left(\frac{1}{x}\right)}\)
The answer is \(0\)
Step 1: \(-1 \le \sin{\left(\frac{1}{x}\right)} \le 1\)
Step 2: \(-|x| \le |x|\sin{\left(\frac{1}{x}\right)} \le |x|\)
Step 3: \(\displaystyle\lim_{x\to 0}(-|x|) \le \displaystyle\lim_{x\to 0}|x|\sin{\left(\frac{1}{x}\right)} \le \displaystyle\lim_{x\to 0}|x|\)
Step 4: \(0 \le \displaystyle\lim_{x\to 0}|x|\sin{\left(\frac{1}{x}\right)} \le 0\)
Step 5: \(\displaystyle\lim_{x\to 0}|x|\sin{\left(\frac{1}{x}\right)}=0\) by Squeeze Theorem
\(\textbf{15)}\) \(\displaystyle\lim_{x\to 0}\frac{x^2\cos{x}}{1+x^2}\)
The answer is \(0\)
Step 1: \(-1 \le \cos{x} \le 1\)
Step 2: \(-x^2 \le x^2\cos{x} \le x^2\)
Step 3: Since \(1+x^2>0\), \(\displaystyle\frac{-x^2}{1+x^2} \le \displaystyle\frac{x^2\cos{x}}{1+x^2} \le \displaystyle\frac{x^2}{1+x^2}\)
Step 4: \(\displaystyle\lim_{x\to 0}\frac{-x^2}{1+x^2} \le \displaystyle\lim_{x\to 0}\frac{x^2\cos{x}}{1+x^2} \le \displaystyle\lim_{x\to 0}\frac{x^2}{1+x^2}\)
Step 5: \(0 \le \displaystyle\lim_{x\to 0}\frac{x^2\cos{x}}{1+x^2} \le 0\)
Step 6: \(\displaystyle\lim_{x\to 0}\frac{x^2\cos{x}}{1+x^2}=0\) by Squeeze Theorem
\(\textbf{16)}\) Suppose \(-x^4 \le g(x) \le x^4\). Find \(\displaystyle\lim_{x\to 0}g(x)\).
The answer is \(0\)
Step 1: \(-x^4 \le g(x) \le x^4\)
Step 2: \(\displaystyle\lim_{x\to 0}(-x^4) \le \displaystyle\lim_{x\to 0}g(x) \le \displaystyle\lim_{x\to 0}x^4\)
Step 3: \(0 \le \displaystyle\lim_{x\to 0}g(x) \le 0\)
Step 4: \(\displaystyle\lim_{x\to 0}g(x)=0\) by Squeeze Theorem
\(\textbf{17)}\) Suppose \(2-x^2 \le f(x) \le 2+x^2\). Find \(\displaystyle\lim_{x\to 0}f(x)\).
The answer is \(2\)
Step 1: \(2-x^2 \le f(x) \le 2+x^2\)
Step 2: \(\displaystyle\lim_{x\to 0}(2-x^2) \le \displaystyle\lim_{x\to 0}f(x) \le \displaystyle\lim_{x\to 0}(2+x^2)\)
Step 3: \(2 \le \displaystyle\lim_{x\to 0}f(x) \le 2\)
Step 4: \(\displaystyle\lim_{x\to 0}f(x)=2\) by Squeeze Theorem
\(\textbf{18)}\) \(\displaystyle\lim_{x\to 0}\left(x^2\sin{\frac{1}{x}}+3\right)\)
The answer is \(3\)
Step 1: \(-1 \le \sin{\frac{1}{x}} \le 1\)
Step 2: \(-x^2 \le x^2\sin{\frac{1}{x}} \le x^2\)
Step 3: \(3-x^2 \le x^2\sin{\frac{1}{x}}+3 \le 3+x^2\)
Step 4: \(\displaystyle\lim_{x\to 0}(3-x^2) \le \displaystyle\lim_{x\to 0}\left(x^2\sin{\frac{1}{x}}+3\right) \le \displaystyle\lim_{x\to 0}(3+x^2)\)
Step 5: \(3 \le \displaystyle\lim_{x\to 0}\left(x^2\sin{\frac{1}{x}}+3\right) \le 3\)
Step 6: \(\displaystyle\lim_{x\to 0}\left(x^2\sin{\frac{1}{x}}+3\right)=3\) by Squeeze Theorem
\(\textbf{19)}\) \(\displaystyle\lim_{x\to \infty}\frac{x\sin{x}}{x^2+1}\)
The answer is \(0\)
Step 1: \(-1 \le \sin{x} \le 1\)
Step 2: \(-x \le x\sin{x} \le x\) for \(x>0\)
Step 3: \(\displaystyle\frac{-x}{x^2+1} \le \displaystyle\frac{x\sin{x}}{x^2+1} \le \frac{x}{x^2+1}\)
Step 4: \(\displaystyle\lim_{x\to \infty}\frac{-x}{x^2+1} \le \displaystyle\lim_{x\to \infty}\frac{x\sin{x}}{x^2+1} \le \displaystyle\lim_{x\to \infty}\frac{x}{x^2+1}\)
Step 5: \(0 \le \displaystyle\lim_{x\to \infty}\frac{x\sin{x}}{x^2+1} \le 0\)
Step 6: \(\displaystyle\lim_{x\to \infty}\frac{x\sin{x}}{x^2+1}=0\) by Squeeze Theorem
\(\textbf{20)}\) \(\displaystyle\lim_{x\to 0}\cos{\left(\frac{1}{x}\right)}\)
The answer is Does Not Exist (DNE)
Step 1: \(-1 \le \cos{\left(\frac{1}{x}\right)} \le 1\)
Step 2: \(\displaystyle\lim_{x\to 0}(-1) \le \displaystyle\lim_{x\to 0}\cos{\left(\frac{1}{x}\right)} \le \displaystyle\lim_{x\to 0}1\)
Step 3: \(-1 \le \displaystyle\lim_{x\to 0}\cos{\left(\frac{1}{x}\right)} \le 1\)
Step 4: The Squeeze Theorem does not prove a limit here since \(-1 \neq 1\). In fact, \(\displaystyle\lim_{x\to 0}\cos{\left(\frac{1}{x}\right)}\) does not exist.
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