Squeeze Theorem Practice

Sample questions

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1. Sample 1difficulty 2/5

   Evaluate $\displaystyle\lim_{x \to 0}\dfrac{\sin(5x)}{x}$.

   • A

     $\tfrac{1}{5}$

   • B

     $1$

   • C

     $0$

   • D

     $5$

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   Why

   Rewrite: $\dfrac{\sin(5x)}{x} = 5 \cdot \dfrac{\sin(5x)}{5x}$. As $x \to 0$, $\sin(5x)/(5x) \to 1$, so the limit is $5$.

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2. Sample 2difficulty 2/5

   Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\tan(3x)}{x}$.

   • A

     $1$

   • B

     $3$

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   • C

     $0$

   • D

     $\tfrac{1}{3}$

   Why

   $\tan(3x)/x = (\sin 3x)/(x\cos 3x) = 3 \cdot (\sin 3x)/(3x) \cdot 1/\cos 3x \to 3 \cdot 1 \cdot 1 = 3$.

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3. Sample 3difficulty 3/5

   Evaluate $\displaystyle\lim_{x \to 0}\dfrac{1 - \cos x}{x}$.

   • A

     $-1$

   • B

     $0$

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   • C

     $\tfrac{1}{2}$

   • D

     $1$

   Why

   Standard limit: $\lim_{x\to 0}(1-\cos x)/x = 0$. (Multiply by conjugate: $\dfrac{1-\cos^2 x}{x(1+\cos x)} = \dfrac{\sin^2 x}{x(1+\cos x)} \to 0 \cdot 1/2 = 0$.)

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4. Sample 4difficulty 3/5

   Evaluate $\displaystyle\lim_{x\to 0} x \sin(1/x)$.

   • A

     $\infty$

   • B

     $0$

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   • C

     $1$

   • D

     Does not exist

   Why

   $|x \sin(1/x)| \le |x|$ since $|\sin(\cdot)| \le 1$. Squeeze theorem: limit is $0$.

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5. Sample 5difficulty 3/5

   Evaluate $\displaystyle\lim_{x\to\infty}\dfrac{\sin x}{x}$.

   • A

     Oscillates — DNE

   • B

     $0$

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   • C

     $1$

   • D

     $\infty$

   Why

   $|\sin x/x| \le 1/|x| \to 0$. Squeeze gives the limit $0$.

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