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Indeterminate Limits---Cosine Forms

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Quick Overview

  • lim
  • The denominator must be the same as the argument of the cosine, and both must be going to zero in the limit.
  • When needed, multiply by the conjugate 1+\cos \theta and use the Pythagorean Identity (see Examples 3 and 4).
  • The cosine forms are often combined with sine forms (so be sure to study those first).

Derivation of the Basic Limit

Show that \displaystyle \lim_{\theta\to0} \frac{1-\cos \theta}\theta = 0

Step 1

Multiply by the conjugate of the numerator.

\begin{align*} \lim_{\theta\to0}\,\frac{1-\cos \theta}\theta & = \lim_{\theta\to0}\,\frac{1-\cos \theta}\theta \cdot \blue{\frac{1+\cos \theta}{1+\cos \theta}}\\[6pt] & = \lim_{\theta\to0}\,\frac{\blue{1-\cos^2 \theta}}{\theta(1+\cos \theta)} \end{align*}

Step 2

Use Pythagorean Identity .

\displaystyle\lim_{\theta\to0}\,\frac{\blue{1-\cos^2 \theta}}{\theta(1+\cos \theta)} = \displaystyle\lim_{\theta\to0}\,\frac{\blue{\sin^2 \theta}}{\theta(1+\cos \theta)}

Step 3

Rewrite to isolate a \frac{\sin \theta} \theta. Then evaluate the limit .

\begin{align*} \lim_{\theta\to0}\,\frac{\red{\sin^2 \theta}}{\red \theta(1+\cos \theta)} & = \lim_{\theta\to0}\left(\red{\frac{\sin \theta} \theta} \cdot \frac{\red{\sin \theta}}{1+\cos \theta}\right)\\[6pt] & = (\red 1) \cdot \frac{\red{\sin 0}}{1+\cos 0}\\[6pt] & = \frac 0 2\\[6pt] & = 0 \end{align*}

Answer

We have shown that

\displaystyle \lim_{\theta\to0} \frac{1-\cos \theta}\theta = 0

Examples

Example 1

Evaluate: \displaystyle \lim_{x\to0}\, \frac{1 - \cos 4x}{x}

Step 1

Multiply by \frac 4 4

\\ \begin{align*} \lim_{x\to0}\, \frac{1 - \cos 4x}{x} % & = \lim_{x\to0}\left(% \blue{\frac 4 4}\cdot \frac{1 - \cos 4x}{x} \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue 4} 1\cdot \frac{1 - \cos 4x}{\blue 4 x} \right)\\[6pt] % & = \blue{4}\cdot\lim_{x\to0}\,\frac{1 - \cos 4x}{4 x} \end{align*} \\

Step 2

Evaluate the limit .

4\cdot\red{\displaystyle\lim_{x\to0}\,\frac{1 - \cos 4x}{4 x}} % = 4(\red 0) % = 0

Answer

\displaystyle \lim_{x\to0}\, \frac{1 - \cos 4x}{x} = 0

Example 2

Evaluate: \displaystyle \lim_{x\to0}\, \frac{1-\cos 2x}{\sin 3x}

Step 1

Rewrite the numerator and denominator as separate fractions.

\displaystyle\lim_{x\to0}\, \frac{1-\cos 2x}{\sin 3x} % = \displaystyle\lim_{x\to0} \left(% \frac{1-\cos 2x} 1 \cdot% \frac 1 {\sin 3x}% \right)

Step 2

Multiply by \frac{2x}{2x} and \frac{3x}{3x}

\begin{align*} \lim_{x\to0} \left(% \frac{1-\cos 2x} 1 \cdot \frac 1 {\sin 3x} \right) % & = \lim_{x\to0} \left(% \blue{\frac{2x}{2x}}\cdot \frac{1-\cos 2x} 1 \cdot \red{\frac{3x}{3x}}\cdot \frac 1 {\sin 3x} \right)\\[6pt] % & = \lim_{x\to0} \left(% \frac{\blue{2x}} 1\cdot \frac{1-\cos 2x}{\blue{2x}}\cdot \frac{1}{\red{3x}}\cdot \frac{\red{3x}}{\sin 3x} \right) \end{align*}

Step 3

Simplify the non-trigonometric fractions.

\begin{align*} \lim_{x\to0} \left(% \blue{\frac{2x} 1}\cdot \frac{1-\cos 2x}{2x}\cdot \red{\frac{1}{3x}}\cdot \frac{3x}{\sin 3x} \right) % & = \lim_{x\to0} \left(% \frac{\blue{2x}}{\red{3x}}\cdot \frac{1-\cos 2x}{2x}\cdot \frac{3x}{\sin 3x} \right)\\[6pt] % & = \lim_{x\to0} \left(% \frac{\blue 2}{\red 3}\cdot \frac{1-\cos 2x}{2x}\cdot \frac{3x}{\sin 3x} \right)\\[6pt] % & = \frac{\blue 2}{\red 3}\cdot\lim_{x\to0}\left(% \frac{1-\cos 2x}{2x}\cdot \frac{3x}{\sin 3x} \right) \end{align*}

Step 4

Evaluate the limit of each factor.

\begin{align*}% \frac 2 3\cdot\lim_{x\to0}\left(% \blue{\frac{1-\cos 2x}{2x}}\cdot \red{\frac{3x}{\sin 3x}} \right) % & =\frac 2 3\,\left(% \blue{\lim_{x\to0}\,\frac{1-\cos 2x}{2x}} \right) % \left(% \red{\lim_{x\to0}\,\frac{3x}{\sin 3x}} \right)\\[6pt] % & = \frac 2 3(\blue 0)(\red 1)\\[6pt] % & = 0 \end{align*}

Answer

\\ \displaystyle \lim_{x\to0}\, \frac{1-\cos 2x}{\sin 3x} = 0 \\

Example 3

Evaluate: \displaystyle \lim_{x\to0} \frac{\sin^2 3x}{1-\cos 2x}

Step 1

Multiply by the conjugate of the denominator.

\begin{align*} \lim_{x\to0} \frac{\sin^2 3x }{1-\cos 2x} % & = \lim_{x\to0}\left(% \frac{\sin^2 3x}{1-\cos 2x }\cdot \blue{\frac{1+\cos 2x}{1+\cos 2x}} \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\sin^2 3x}{\blue{1-\cos^2 2x}}\cdot \frac{\blue{1+\cos 2x}} 1 \right) \end{align*}

Step 2

Use Pythagorean Identity to convert the denominator to sines.

\displaystyle\lim_{x\to0}\left(% \frac{\sin^2 3x}{\blue{1-\cos^2 2x}}\cdot \frac{1+\cos 2x} 1 \right) % = \displaystyle\lim_{x\to0} \left(% \frac{\sin^2 3x}{\blue{\sin^2 2x}}\cdot \frac{1+\cos 2x} 1 \right)

Step 3

Separate each term into its own fraction.

\displaystyle\lim_{x\to0} \left(% \frac{\red{\sin^2 3x}}{\blue{\sin^2 2x}}\cdot \frac{1+\cos 2x} 1 \right) % = \displaystyle\lim_{x\to0}\left(% \red{\frac{\sin 3x} 1} \cdot \red{\frac{\sin 3x} 1} \cdot \blue{\frac 1 {\sin 2x}} \cdot \blue{\frac 1 {\sin 2x}} \cdot \frac{1+\cos 2x} 1 \right)

Step 4

For each fraction with a sine function, multiply by \frac{2x}{2x} or \frac{3x}{3x} as appropriate.

\begin{align*} & \lim_{x\to0}\left(% \frac{\sin 3x } 1 \cdot \frac{\sin 3x} 1 \cdot \frac 1 {\sin 2x} \cdot \frac 1 {\sin 2x } \cdot \frac{1+\cos 2x} 1 \right)\\[6pt] % & = \lim_{x\to0}\left(% \blue{\frac{3x}{3x}}\cdot\frac{\sin 3x} 1 \cdot \blue{\frac{3x}{3x}}\cdot\frac{\sin 3x} 1 \cdot \red{\frac{2x}{2x}}\cdot\frac 1 {\sin 2x} \cdot \red{\frac{2x}{2x}}\cdot\frac 1 {\sin 2x} \cdot \frac{1+\cos 2x} 1 \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{3x}} 1 \cdot\frac{\sin 3x}{\blue{3x}}\cdot \frac{\blue{3x}} 1 \cdot\frac{\sin 3x}{\blue{3x}} \cdot \frac 1 {\red{2x}} \cdot\frac{\red{2x}}{\sin 2x} \cdot \frac 1 {\red{2x}} \cdot\frac{\red{2x}}{\sin 2x} \cdot \frac{1+\cos 2x} 1 \right) \end{align*}

Step 5

Simplify the non-trigonometric fractions.

\begin{align*} & \quad\lim_{x\to0}\left(% \blue{\frac{3x} 1} \cdot\frac{\sin 3x}{3x}\cdot \blue{\frac{3x} 1} \cdot\frac{\sin 3x}{3x} \cdot \red{\frac 1 {2x}} \cdot\frac{2x}{\sin 2x} \cdot \red{\frac 1 {2x}} \cdot\frac{2x}{\sin 2x} \cdot \frac{1+\cos 2x} 1 \right)\\[6pt] % & = \lim_{x\to0}\left(% \blue{\frac{3x} 1} \cdot \blue{\frac{3x} 1} \cdot \red{\frac 1 {2x}} \cdot \red{\frac 1 {2x}} \cdot \frac{\sin 3x}{3x}\cdot \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{2x}{\sin 2x} \cdot \frac{1+\cos 2x} 1 \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{9x^2}}{\red{4x^2}} \cdot \frac{\sin 3x}{3x}\cdot \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{2x}{\sin 2x} \cdot \frac{1+\cos 2x} 1 \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{9}}{\red{4}} \cdot \frac{\sin 3x}{3x}\cdot \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{2x}{\sin 2x} \cdot \frac{1+\cos 2x} 1 \right)\\[6pt] % & = \frac{\blue{9}}{\red{4}}\cdot\lim_{x\to0}\left(% \frac{\sin 3x}{3x}\cdot \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{2x}{\sin 2x} \cdot \frac{1+\cos 2x} 1 \right) \end{align*}

Step 6

Evaluate the limit of each factor.

\\ \begin{align*} & \frac 9 4\cdot\lim_{x\to0}\left(% \blue{\frac{\sin 3x}{3x}}\cdot \blue{\frac{\sin 3x}{3x}} \cdot \red{\frac{2x}{\sin 2x}} \cdot \red{\frac{2x}{\sin 2x}} \cdot \frac{1+\cos 2x} 1 \right)\\[6pt] % & = \frac 9 4(\blue 1)(\blue 1)(\red 1)(\red 1)(1 + \cos 0)\\[6pt] % & = \frac 9 4 \cdot 2\\[6pt] % & = \frac 9 2 \end{align*} \\

Answer

\displaystyle \lim_{x\to0}\, \frac{\sin^2 3x}{1-\cos 2x} = \frac 9 2

Example 4

Evaluate: \displaystyle\lim_{x\to0}\,\frac{1-\cos x}{\cos 6x -1}

Step 1

Write the numerator and denominator in separate fractions.

\displaystyle\lim_{x\to0}\,\frac{1-\cos x}{\cos 6x -1} % = \displaystyle\lim_{x\to0}\left(% \frac{1-\cos x} 1 \cdot \frac 1 {\cos 6x -1} \right)

Step 2

For each fraction, use the conjugate to get a difference of squares.

\begin{align*} \lim_{x\to0}\,\left(\frac{1-\cos x} 1 \cdot \frac 1 {\cos 6x -1}\right) % & = \lim_{x\to0}\left(% \frac{1-\cos x} 1\cdot \frac{\blue{1+\cos x}}{\blue{1+\cos x}}\cdot \frac 1 {\cos 6x -1}\cdot \frac{\red{\cos 6x + 1}}{\red{\cos 6x + 1}} \right)\\[6pt] % & \lim_{x\to0}\left(% \frac{\blue{1-\cos^2 x}}{1+\cos x}\cdot \frac{\cos 6x + 1}{\red{\cos^2 6x -1}} \right) \end{align*}

Step 3

Use Pythagorean Identity to change the difference of squares into sines.

\begin{align*}% \lim_{x\to0}\left(% \frac{\blue{1-\cos^2 x}}{1+\cos x}\cdot \frac{\cos 6x + 1}{\red{\cos^2 6x -1}} \right) % & = \lim_{x\to0}\left(% \frac{\blue{\sin^2 x}}{1+\cos x}\cdot \frac{\cos 6x + 1}{\red{-\sin^2 6x}} \right)\\[6pt] % & = \red{-1}\cdot\lim_{x\to0}\left(% \frac{\sin^2 x}{1+\cos x}\cdot \frac{\cos 6x + 1}{\sin^2 6x} \right) \end{align*}

Step 4

Separate the sines into their own fractions.

\begin{align*}% &\quad-1\cdot\lim_{x\to0}\left(% \frac{\blue{\sin^2 x}}{1+\cos x}\cdot \frac{\cos 6x + 1}{\red{\sin^2 6x}} \right)\\[6pt] % & = -1\cdot\lim_{x\to0}\left(% \blue{\frac{\sin x} 1} \cdot \blue{\frac{\sin x} 1} \cdot \red{\frac 1 {\sin 6x}} \cdot \red{\frac 1 {\sin 6x}} \cdot \frac{\cos 6x +1}{1+\cos x} \right) \end{align*}

Step 5

Multiply each of the sine fractions by \frac x x or \frac{6x}{6x} as appropriate.

\begin{align*} & -1\cdot\lim_{x\to0}\left(% \frac{\sin x} 1 \cdot \frac{\sin x} 1 \cdot \frac 1 {\sin 6x} \cdot \frac 1 {\sin 6x} \cdot \frac{\cos 6x +1}{1+\cos x} \right)\\[6pt] % & = -1\cdot\lim_{x\to0}\left(% \blue{\frac x x}\cdot \frac{\sin x} 1 \cdot \blue{\frac x x}\cdot \frac{\sin x} 1 \cdot \red{\frac{6x}{6x}}\cdot\frac 1 {\sin 6x} \cdot \red{\frac{6x}{6x}}\cdot\frac 1 {\sin 6x} \cdot \frac{\cos 6x +1}{1+\cos x} \right)\\[6pt] % & = -1\cdot\lim_{x\to0}\left(% \frac{\blue x} 1\cdot \frac{\sin x}{\blue x} \cdot \frac{\blue x} 1\cdot \frac{\sin x}{\blue x} \cdot \frac 1 {\red {6x}}\cdot\frac{\red{6x}}{\sin 6x} \cdot \frac 1 {\red {6x}}\cdot\frac{\red{6x}}{\sin 6x} \cdot \frac{\cos 6x +1}{1+\cos x} \right) \end{align*}

Step 6

Simplify the non-trigonometric fractions.

\\ \begin{align*} & -1\cdot\lim_{x\to0}\left(% \blue{\frac x 1}\cdot \frac{\sin x} x \cdot \blue{\frac x 1}\cdot \frac{\sin x} x \cdot \red{\frac 1 {6x}}\cdot\frac{6x}{\sin 6x} \cdot \red{\frac 1 {6x}}\cdot\frac{6x}{\sin 6x} \cdot \frac{\cos 6x +1}{1+\cos x} \right)\\[6pt] % & = -1\cdot\lim_{x\to0}\left(% \blue{\frac x 1}\cdot \blue{\frac x 1}\cdot \red{\frac 1 {6x}}\cdot\red{\frac 1 {6x}}\cdot \frac{\sin x} x \cdot \frac{\sin x} x \cdot \frac{6x}{\sin 6x} \cdot \frac{6x}{\sin 6x} \cdot \frac{\cos 6x +1}{1+\cos x} \right)\\[6pt] % & = -1\cdot\lim_{x\to0}\left(% \frac{\blue{x^2}}{\red{36x^2}}\cdot \frac{\sin x} x \cdot \frac{\sin x} x \cdot \frac{6x}{\sin 6x} \cdot \frac{6x}{\sin 6x} \cdot \frac{\cos 6x +1}{1+\cos x} \right)\\[6pt] % & = -1\cdot\lim_{x\to0}\left(% \frac{\blue{1}}{\red{36}}\cdot \frac{\sin x} x \cdot \frac{\sin x} x \cdot \frac{6x}{\sin 6x} \cdot \frac{6x}{\sin 6x} \cdot \frac{\cos 6x +1}{1+\cos x} \right)\\[6pt] % & = -\frac{\blue{1}}{\red{36}}\cdot\lim_{x\to0}\left(% \frac{\sin x} x \cdot \frac{\sin x} x \cdot \frac{6x}{\sin 6x} \cdot \frac{6x}{\sin 6x} \cdot \frac{\cos 6x +1}{1+\cos x} \right) \end{align*} \\

Step 7

Evaluate the limit of each factor.

\\ \begin{align*} & -\frac 1{36}\cdot\lim_{x\to0}\left( \blue{\frac{\sin x} x} \cdot \blue{\frac{\sin x} x} \cdot \red{\frac{6x}{\sin 6x}} \cdot \red{\frac{6x}{\sin 6x}} \cdot \frac{\cos 6x +1}{1+\cos x} \right)\\[6pt] % & = -\frac 1 {36} (\blue 1)(\blue 1)(\red 1)(\red 1)\left(\frac{\cos 0 +1}{1 + \cos 0}\right)\\[6pt] % & = -\frac 1 {36}\left(\frac 2 2\right)\\[6pt] % & = -\frac 1 {36} \end{align*} \\

Answer

\displaystyle\lim_{x\to0}\,\frac{1-\cos x}{\cos 6x -1} = -\frac 1 {36}

Practice Problems

Problem 1

Evaluate: \displaystyle \lim_{x\to0} \,\frac{1-\cos 7x}{2x}

Problem 2

Evaluate: \displaystyle \lim_{x\to0}\,\frac{\cos 3x - 1}{4x}

Problem 3

Evaluate: \displaystyle \lim_{x\to0}\,\frac{8x}{1-\cos x}

Problem 4

Evaluate: \displaystyle \lim_{x\to0}\,\frac{x^2}{1-\cos 12x}

Problem 5

Evaluate: \displaystyle \lim_{x\to0}\,\frac{1-\cos 9x}{\sin 4x}

Problem 6

Evaluate: \displaystyle \lim_{x\to0}\,\frac{1-\cos 2x}{x\sin 4x}

Problem 7

Evaluate: \displaystyle \lim_{x\to0}\,\frac{x\sin 3x}{1-\cos 7x}

Problem 8

Evaluate: \displaystyle \lim_{x\to0}\,\frac{1-\cos 4x}{1 - \cos 8x}

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