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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
Step 1
Factor out the 2 from the denominator and multiply by $$\frac 7 7$$.
$$
\begin{align*}
\lim_{x\to0}\, \frac{1-\cos 7x}{\blue 2 x}
%
& = \lim_{x\to0}\left(%
\frac 1 {\blue 2}\cdot
\frac{1-\cos 7x}{x}
\right)\\[6pt]
%
& = \frac 1 {\blue 2}\cdot\lim_{x\to0} \left(%
\red{\frac 7 7}\cdot
\frac{1-\cos 7x}{x}
\right)\\[6pt]
%
& = \frac 1 2\cdot\lim_{x\to0}\left(%
\frac{\red 7} 1\cdot
\frac{1-\cos 7x}{\red 7 x}
\right)\\[6pt]
%
& = \frac{\red 7} 2\cdot\lim_{x\to0}\,\frac{1-\cos 7x}{\red 7 x}
\end{align*}
$$
Step 2
$$
\frac 7 2\cdot\blue{\displaystyle\lim_{x\to0} \, \frac{1-\cos 7x}{7x}}
%
= \frac 7 2 (\blue 0)
%
= 0
$$
Answer
$$\displaystyle \lim_{x\to0}\, \frac{1-\cos 7x}{2x} = 0$$.
Step 1
Factor out a negative 1 from the numerator and the 4 from the denominator.
$$
\begin{align*}%
\lim_{x\to0}\,\frac{\cos 3x - 1}{4x}
%
& = \lim_{x\to0}\,\frac{\blue{-1}(1 - \cos 3x)}{\red 4 x}\\[6pt]
%
& = \lim_{x\to0}\,\left(%
\blue{-\frac 1 {\red 4}} \cdot
\frac{1 - \cos 3x}{x}
\right)\\[6pt]
%
& = \blue{-\frac 1 {\red 4}} \cdot\lim_{x\to0}\,\frac{1-\cos 3x} x
\end{align*}
$$
Step 2
Multiply by $$\frac 3 3$$.
$$
\begin{align*}
-\frac 1 4 \cdot\lim_{x\to0}\,\frac{1-\cos 3x} x
%
& = -\frac 1 4 \cdot\lim_{x\to0}\left(%
\frac{\blue 3}{\blue 3}\cdot
\frac{1-\cos 3x} x
\right)\\[6pt]
%
& = -\frac 1 4 \cdot\lim_{x\to0}\left(%
\frac{\blue 3} 1\cdot
\frac{1-\cos 3x}{\blue 3 x}
\right)\\[6pt]
%
& = -\frac{\blue 3} 4 \cdot\lim_{x\to0}\, \frac{1-\cos 3x}{\blue 3 x}
\end{align*}
$$
Step 3
$$
-\frac 3 4 \cdot\blue{\displaystyle\lim_{x\to0}\, \frac{1-\cos 3x}{3 x}}
%
= -\frac 3 4 (\blue 0)
%
= 0
$$
Answer
$$\displaystyle \lim_{x\to0}\,\frac{\cos 3x - 1}{4x} = 0$$.
Step 1
Factor the 8 out of the numerator.
$$
\displaystyle\lim_{x\to0}\,\frac{\blue 8 x}{1-\cos x}
%
= \blue 8\cdot \displaystyle\lim_{x\to0}\,\frac{x}{1-\cos x}
$$
Step 2
Multiply by the conjugate of the denominator to get a difference of squares.
$$
\begin{align*}
8\cdot \lim_{x\to0}\,\frac{x}{1-\cos x}
%
& = 8\cdot \lim_{x\to0}\left(%
\frac{x}{1-\cos x} \cdot
\blue{\frac{1+\cos x}{1+\cos x}}
\right)\\[6pt]
%
& = 8\cdot \lim_{x\to0}\left(%
\frac{x}{\blue{1-\cos^2 x}}\cdot
\frac{\blue{1+\cos x}} 1
\right)
\end{align*}
$$
Step 3
$$
8\cdot \displaystyle\lim_{x\to0}\left(%
\frac{x}{\blue{1-\cos^2 x}}\cdot
\frac{1+\cos x} 1
\right)
%
= 8\cdot \displaystyle\lim_{x\to0}\left(%
\frac{x}{\blue{\sin^2 x}}\cdot
\frac{1+\cos x} 1
\right)
$$
Step 4
Separate into the sine functions.
$$
8\cdot \displaystyle\lim_{x\to0}\left(%
\frac{x}{\blue{\sin^2 x}}\cdot
\frac{1+\cos x} 1
\right)
%
= 8\cdot \displaystyle\lim_{x\to0}\left(%
\frac x {\blue{\sin x}} \cdot
\frac{1+\cos x}{\blue{\sin x}}
\right)
$$
Step 5
Evaluate the limit of each factor.
$$
8\cdot \displaystyle\lim_{x\to0}\left(%
\blue{\frac x {\sin x}} \cdot
\red{\frac{1+\cos x}{\sin x}}
\right)
%
= 8 (\blue 1)\left(\red{\frac 2 0}\right)
%
= \frac{16} 0\qquad \mbox{Division by zero!}
$$
Answer
$$\displaystyle \lim_{x\to0}\,\frac{8x}{1-\cos x}$$ does not exist.
Step 1
Multiply by the conjugate of the denominator to get a difference of squares.
$$
\begin{align*}
\lim_{x\to0}\,\frac{x^2}{1-\cos 12x}
%
& = \lim_{x\to0}\left(%
\frac{x^2}{1-\cos 12x}\cdot
\blue{\frac{1+\cos 12x}{1+\cos 12x}}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{x^2}{\blue{1-\cos^2 12x}}\cdot
\frac{\blue{1+\cos 12x}} 1
\right)
\end{align*}
$$
Step 2
$$
\displaystyle\lim_{x\to0}\left(%
\frac{x^2}{\blue{1-\cos^2 12x}}\cdot
\frac{1+\cos 12x} 1
\right)
%
= \displaystyle\lim_{x\to0}\left(%
\frac{x^2}{\blue{\sin^2 12x}}\cdot
\frac{1+\cos 12x} 1
\right)
$$
Step 3
Rewrite as separate fractions.
$$
\displaystyle\lim_{x\to0}\left(%
\frac{x^2}{\sin^2 12x}\cdot
\frac{1+\cos 12x} 1
\right)
%
=
\displaystyle\lim_{x\to0}\left(%
\frac x {\sin 12x} \cdot
\frac x {\sin 12 x}\cdot
\frac{1+\cos 12x}{1}
\right)
$$
Step 4
Multiply by $$\frac{12}{12}$$ for each sine fraction.
$$
\begin{align*}
&\quad \lim_{x\to0}\left(%
\frac x {\sin 12x} \cdot
\frac x {\sin 12 x}\cdot
\frac{1+\cos 12x}{1}\right)
\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{12}}{\blue{12}}\cdot
\frac x {\sin 12x} \cdot
\frac{\blue{12}}{\blue{12}}\cdot
\frac x {\sin 12 x}\cdot
\frac{1+\cos 12x}{1}\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{1}{\blue{12}}\cdot
\frac{\blue{12} x} {\sin 12x} \cdot
\frac{1}{\blue{12}}\cdot
\frac {\blue{12}x} {\sin 12 x}\cdot
\frac{1+\cos 12x}{1}
\right)
\end{align*}
$$
Step 5
$$
\begin{align*}%
&\quad \lim_{x\to0}\left(%
\blue{\frac 1 {12}}\cdot
\frac{12 x} {\sin 12x} \cdot
\blue{\frac 1 {12}}\cdot
\frac {12x} {\sin 12 x}\cdot
\frac{1+\cos 12x}{1}
\right)
\\[6pt]
%
& = \lim_{x\to0}\left(%
\blue{\frac 1 {144}}\cdot
\frac{12 x} {\sin 12x} \cdot
\frac {12x} {\sin 12 x}\cdot
\frac{1+\cos 12x}{1}
\right)\\[6pt]
%
& = \blue{\frac 1 {144}}\cdot\lim_{x\to0}\left(%
\frac{12 x} {\sin 12x} \cdot
\frac {12x} {\sin 12 x}\cdot
\frac{1+\cos 12x}{1}
\right)
\end{align*}
$$
Step 6
Evaluate the limit of each factor.
$$
\frac 1 {144}\cdot\displaystyle\lim_{x\to0}\left(%
\blue{\frac{12 x} {\sin 12x}} \cdot
\red{\frac {12x} {\sin 12 x}}\cdot
\frac{1+\cos 12x}{1}
\right)
%
= \frac 1 {144} (\blue 1)(\red 1)(1 + \cos 0)
%
= \frac 2 {144}
%
= \frac 1 {72}
$$
Answer
$$
\displaystyle \lim_{x\to0}\,\frac{x^2}{1-\cos 12x} = \frac 1 {72}
$$
Step 1
Rewrite as separate fractions.
$$
\displaystyle\lim_{x\to0}\,\frac{1-\cos 9x}{\sin 4x}
%
= \displaystyle\lim_{x\to0}\left(%
\frac{1-\cos 9x} 1 \cdot
\frac 1 {\sin 4x}
\right)
$$
Step 2
Multiply by $$\frac{9x}{9x}$$ and $$\frac{4x}{4x}$$.
$$
\begin{align*}
\lim_{x\to0}\left(%
\frac{1-\cos 9x} 1 \cdot
\frac 1 {\sin 4x}
\right)
%
& = \lim_{x\to0}\left(%
\frac{\blue{9x}}{\blue{9x}}\cdot
\frac{1-\cos 9x} 1 \cdot
\frac{\red{4x}}{\red{4x}}\cdot
\frac 1 {\sin 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{9x}}{1}\cdot
\frac{1-\cos 9x}{\blue{9x}}\cdot
\frac{1}{\red{4x}}\cdot
\frac{\red{4x}} {\sin 4x}
\right)
\end{align*}
$$
Step 3
Simplify the non-trigonometric fractions.
$$
\begin{align*}
\lim_{x\to0}\left(%
\blue{\frac{9x}{1}}\cdot
\frac{1-\cos 9x}{9x} \cdot
\red{\frac{1}{4x}}\cdot
\frac{4x} {\sin 4x}
\right)
%
& = \lim_{x\to0}\left(%
\frac{\blue{9x}}{\red{4x}}\cdot
\frac{1-\cos 9x}{9x}\cdot
\frac{4x} {\sin 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{9}}{\red{4}}\cdot
\frac{1-\cos 9x}{9x}\cdot
\frac{4x} {\sin 4x}
\right)\\[6pt]
%
& = \frac{\blue{9}}{\red{4}}\cdot\lim_{x\to0}\left(%
\frac{1-\cos 9x}{9x}\cdot
\frac{4x} {\sin 4x}
\right)
\end{align*}
$$
Step 4
Evaluate the limit of each factor.
$$
\frac{9}{4}\cdot\displaystyle\lim_{x\to0}\left(%
\blue{\frac{1-\cos 9x}{9x}}\cdot
\red{\frac{4x} {\sin 4x}}
\right)
%
= \frac 9 4 (\blue 0)(\red 1)
%
= 0
$$
Answer
$$
\displaystyle \lim_{x\to0}\,\frac{1-\cos 9x}{\sin 4x} = 0
$$
Step 1
Write as separate fractions.
$$
\displaystyle\lim_{x\to0}\,\frac{1-\cos 2x}{x\sin 4x}
%
= \displaystyle\lim_{x\to0}\left(%
\frac{1-\cos 2x} x \cdot
\frac 1 {\sin 4x}
\right)
$$
Step 2
Multiply by $$\frac 2 2$$ and $$\frac{4x}{4x}$$.
$$
\begin{align*}
\lim_{x\to0}\left(%
\frac{1-\cos 2x} x \cdot
\frac 1 {\sin 4x}
\right)
%
& = \lim_{x\to0}\left(%
\frac{\blue 2}{\blue 2}\cdot
\frac{1-\cos 2x} x \cdot
\frac{\red{4x}}{\red{4x}}\cdot
\frac 1 {\sin 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue 2}{1}\cdot
\frac{1-\cos 2x}{\blue 2 x}\cdot
\frac{1}{\red{4x}}\cdot
\frac{\red{4x}}{\sin 4x}
\right)
\end{align*}
$$
Step 3
Factor out the coefficients.
$$
\begin{align*}
\lim_{x\to0}\left(%
\blue{\frac 2 1}\cdot
\frac{1-\cos 2x}{2x}\cdot
\frac{1}{\red{4}x}\cdot
\frac{4x}{\sin 4x}
\right)
%
& = \lim_{x\to0}\left(%
\frac{\blue 2}{\red 4}\cdot
\frac{1-\cos 2x}{2x}\cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}
\right)\\[6pt]
%
& = \frac{\blue 1}{\red 2}\cdot\lim_{x\to0}\left(%
\frac{1-\cos 2x}{2x}\cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}
\right)
\end{align*}
$$
Step 4
Evaluate the limit of each factor.
$$
\frac 1 2 \cdot\displaystyle\lim_{x\to0}\left(%
\blue{\frac{1-\cos 2x}{2x}}\cdot
\frac{1}{x}\cdot
\red{\frac{4x}{\sin 4x}}
\right)
%
= \frac 1 2 (\blue 0)\left(\frac 1 0\right)(\red 1)
%
= \frac 0 0
$$
This is an indeterminate form. We'll need to use the conjugate $$1 + \cos 2x$$.
Step 5
Multiply by the conjugate $$1+\cos 2x$$.
$$
\begin{align*}
&\quad\frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{1-\cos 2x}{2x} \cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}
\right)
\\[6pt]
%
& = \frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{1-\cos 2x}{2x} \cdot
\blue{\frac{1+\cos 2x}{1+\cos 2x}}\cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}
\right)\\[6pt]
%
& = \frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\blue{1-\cos^2 2x}}{2x} \cdot
\frac{1}{\blue{1+\cos 2x}}\cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}
\right)
\end{align*}
$$
Step 6
$$
\begin{align*}%
&\quad\frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\blue{1-\cos^2 2x}}{2x}\cdot
\frac{1}{1+\cos 2x}\cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}
\right)\\[6pt]
%
& = \frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\blue{\sin^2 2x}}{2x}\cdot
\frac{1}{1+\cos 2x}\cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}\right)
\end{align*}
$$
Step 7
Adjust the fractions to get $$\frac{\sin\theta} \theta$$ forms.
$$
\begin{align*}
&=\quad\frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\blue{\sin^2 2x}}{\blue{2x}}\cdot
\frac{1}{1+\cos 2x}\cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}
\right)
\\[6pt]
&=\frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\blue{\sin 2x}}{\blue{2x}}\cdot
\frac{\blue{\sin 2x}} {\blue 1} \cdot
\frac{1}{1+\cos 2x}\cdot
\frac{1}{x}\cdot
\frac{4x}{\sin 4x}
\right)
\\[6pt]
%
& = \frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\sin 2x}{2x}\cdot
\frac{\red{\sin 2x}} 1 \cdot
\frac{1}{1+\cos 2x}\cdot
\frac{1}{\red x}\cdot
\frac{4x}{\sin 4x}
\right)\\[6pt]
%
& = \frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\sin 2x}{2x}\cdot
\frac{\red{\sin 2x}}{\red x} \cdot
\frac{1}{1+\cos 2x}\cdot
\frac{4x}{\sin 4x}
\right)\\[6pt]
%
& = \frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\sin 2x}{2x}\cdot
\frac{\red 2}{\red 2}\cdot
\frac{\red{\sin 2x}}{\red x} \cdot
\frac{1}{1+\cos 2x}\cdot
\frac{4x}{\sin 4x}
\right)\\[6pt]
%
& = \frac 1 2 \cdot\lim_{x\to0}\left(%
\frac{\sin 2x}{2x}\cdot
\frac{\red 2}{1}\cdot
\frac{\red{\sin 2x}}{\red {2x}}\cdot
\frac{1}{1+\cos 2x}\cdot
\frac{4x}{\sin 4x}
\right)\\[6pt]
%
& = \frac{\red 2} 2 \cdot\lim_{x\to0}\left(%
\frac{\sin 2x}{2x}\cdot
\frac{\red{\sin 2x}}{\red {2x}} \cdot
\frac{1}{1+\cos 2x}\cdot
\frac{4x}{\sin 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\sin 2x}{2x}\cdot
\frac{\sin 2x}{2x} \cdot
\frac{1}{1+\cos 2x}\cdot
\frac{4x}{\sin 4x}
\right)
\end{align*}
$$
Step 8
Evaluate the limit of each factor.
$$
\displaystyle\lim_{x\to0}\left(%
\frac{\sin 2x}{2x}\cdot
\frac{\sin 2x}{2x} \cdot
\frac{1}{1+\cos 2x}\cdot
\frac{4x}{\sin 4x}
\right)
%
= (1)(1)\left(\frac 1 {1 + \cos 0}\right)(1)
%
= \frac 1 2
$$
Answer
$$
\displaystyle \lim_{x\to0}\,\frac{1-\cos 2x}{x\sin 4x} = \frac 1 2
$$
Step 1
Multiply by the conjugate of the denominator.
$$
\begin{align*}
\lim_{x\to0}\,\frac{x\sin 3x}{1-\cos 7x}
%
& = \lim_{x\to0}\left(%
\frac{x\sin 3x}{1-\cos 7x}\cdot
\frac{\blue{1+\cos 7x}}{\blue{1 + \cos 7x}}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{x\sin 3x}{\blue{1-\cos^2 7x}}\cdot
\frac{\blue{1+\cos 7x}}{1}
\right)
\end{align*}
$$
Step 2
$$
\displaystyle\lim_{x\to0}\left(%
\frac{x\sin 3x}{\blue{1-\cos^2 7x}}\cdot
\frac{1+\cos 7x}{1}
\right)
%
= \displaystyle\lim_{x\to0}\left(%
\frac{x\sin 3x}{\blue{\sin^2 7x}}\cdot
\frac{1+\cos 7x}{1}
\right)
$$
Step 3
Rewrite as separate fractions.
$$
\displaystyle\lim_{x\to0}\left(%
\frac{\blue x\sin 3x}{\blue{\sin^2 7x}}\cdot
\frac{1+\cos 7x}{1}
\right)
%
=
\displaystyle\lim_{x\to0}\left(%
\frac{\blue x}{\blue{\sin 7x}}\cdot
\frac 1 {\blue{\sin 7x}}\cdot
\frac{\sin 3x} 1 \cdot
\frac{1+\cos 7x}{1}
\right)
$$.
Step 4
Multiply each sine fraction by the appropriate value to get the correct forms.
$$
\begin{align*}
& \quad\lim_{x\to0}\left(%
\frac x {\sin 7x}\cdot
\frac 1 {\sin 7x}\cdot
\frac{\sin 3x} 1 \cdot
\frac{1+\cos 7x}{1}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue 7}{\blue 7} \cdot
\frac x {\sin 7x} \cdot
\frac{\blue{7x}}{\blue{7x}}\cdot
\frac 1 {\sin 7x} \cdot
\frac{\red{3x}}{\red{3x}}\cdot
\frac{\sin 3x} 1 \cdot
\frac{1+\cos 7x}{1}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{1}{\blue 7} \cdot
\frac{\blue 7 x}{\sin 7x} \cdot
\frac{1}{\blue{7x}}\cdot
\frac{\blue{7x}}{\sin 7x} \cdot
\frac{\red{3x}}{1}\cdot
\frac{\sin 3x}{\red{3x}} \cdot
\frac{1+\cos 7x}{1}\right)
\end{align*}
$$
Step 5
Simplify the non-trigonometric fractions.
$$
\begin{align*}
& \lim_{x\to0}\left(%
\frac{\blue 1}{\blue 7}\cdot
\frac{7 x}{\sin 7x} \cdot
\frac{\blue 1}{\blue{7x}}\cdot
\frac{7x}{\sin 7x} \cdot
\frac{\red{3x}}{\red 1}\cdot
\frac{\sin 3x}{3x} \cdot
\frac{1+\cos 7x}{1}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\red{3x}}{\blue{49x}}\cdot
\frac{7 x}{\sin 7x} \cdot
\frac{7x}{\sin 7x} \cdot
\frac{\sin 3x}{3x} \cdot
\frac{1+\cos 7x}{1}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\red{3}}{\blue{49}}\cdot
\frac{7 x}{\sin 7x} \cdot
\frac{7x}{\sin 7x} \cdot
\frac{\sin 3x}{3x} \cdot
\frac{1+\cos 7x}{1}
\right)\\[6pt]
%
& = \frac{\red{3}}{\blue{49}}\cdot\lim_{x\to0}\left(%
\frac{7 x}{\sin 7x} \cdot
\frac{7x}{\sin 7x} \cdot
\frac{\sin 3x}{3x} \cdot
\frac{1+\cos 7x}{1}
\right)
\end{align*}
$$
Step 6
Evaluate the limit of each factor.
$$
\frac 3 {49} \cdot\displaystyle\lim_{x\to0}\left(%
\blue{\frac{7 x}{\sin 7x}} \cdot
\blue{\frac{7x}{\sin 7x}} \cdot
\red{\frac{\sin 3x}{3x}} \cdot
\frac{1+\cos 7x}{1}\right)
%
= \frac 3 {49}(\blue 1)(\blue 1)(\red 1)(1 + \cos 0)
%
= \frac 6 {49}
$$
Answer
$$
\displaystyle \lim_{x\to0}\,\frac{x\sin 3x}{1-\cos 7x} = \frac 6 {49}
$$
Step 1
Rewrite as separate fractions.
$$
\displaystyle\lim_{x\to0}\,\frac{1-\cos 4x}{1 - \cos 8x}
%
= \displaystyle\lim_{x\to0}
\left(%
\frac{1-\cos 4x} 1 \cdot
\frac 1 {1 - \cos 8x}
\right)
$$
Step 2
Multiply by the conjugates to get difference of squares.
$$
\begin{align*}
&\quad\lim_{x\to0}\left(%
\frac{1-\cos 4x} 1 \cdot
\frac 1 {1 - \cos 8x}
\right)
\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{1+\cos 4x}}{\blue{1+\cos 4x}}\cdot
\frac{1-\cos 4x} 1 \cdot
\frac 1 {1 - \cos 8x}\cdot
\frac{\red{1+\cos 8x}}{\red{1+\cos 8x}}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{1-\cos^2 4x}}{1+\cos 4x}\cdot
\frac{1+\cos 8x}{\red{1-\cos^2 8x}}
\right)
\end{align*}
$$
Step 3
$$
\displaystyle\lim_{x\to0}\left(%
\frac{\blue{1-\cos^2 4x}}{1+\cos 4x} \cdot
\frac{1+\cos 8x}{\red{1-\cos^2 8x}}
\right)
%
= \displaystyle\lim_{x\to0}\left(%
\frac{\blue{\sin^2 4x}}{1+\cos 4x} \cdot
\frac{1+\cos 8x}{\red{\sin^2 8x}}
\right)
$$.
Step 4
Write so each sine is in a separate fraction.
$$
\\
\displaystyle\lim_{x\to0}\left(%
\frac{\blue{\sin^2 4x}}{1+\cos 4x} \cdot
\frac{1+\cos 8x}{\red{\sin^2 8x}}
\right)
%
= \displaystyle\lim_{x\to0}\left(%
\frac{\blue{\sin 4x}} 1 \cdot
\frac{\blue{\sin 4x}} 1 \cdot
\frac 1 {\red{\sin 8x}} \cdot
\frac 1 {\red{\sin 8x}} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)
\\
$$
Step 5
Multiply by $$\frac{4x}{4x}$$ and $$\frac{8x}{8x}$$ as appropriate.
$$
\begin{align*}
& \quad\lim_{x\to0}\left(%
\frac{\sin 4x} 1 \cdot
\frac{\sin 4x} 1 \cdot
\frac 1 {\sin 8x} \cdot
\frac 1 {\sin 8x} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{4x}}{\blue{4x}}\cdot
\frac{\sin 4x} 1 \cdot
\frac{\blue{4x}}{\blue{4x}}\cdot
\frac{\sin 4x} 1 \cdot
\frac{\red{8x}}{\red{8x}}\cdot
\frac 1 {\sin 8x} \cdot
\frac{\red{8x}}{\red{8x}}\cdot
\frac 1 {\sin 8x} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{4x}}{1}\cdot
\frac{\sin 4x}{\blue{4x}} \cdot
\frac{\blue{4x}} 1\cdot
\frac{\sin 4x}{\blue{4x}} \cdot
\frac 1 {\red{8x}}\cdot
\frac{\red{8x}}{\sin 8x} \cdot
\frac 1 {\red{8x}}\cdot
\frac{\red{8x}}{\sin 8x} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)
\end{align*}
$$
Step 6
Simplify the non-trigonometric fractions.
$$
\begin{align*}
& \quad\lim_{x\to0}\left(%
\frac{\blue{4x}}{\blue 1}\cdot
\frac{\sin 4x}{4x} \cdot
\frac{\blue{4x}}{\blue1}\cdot
\frac{\sin 4x}{4x} \cdot
\frac{\red 1}{\red{8x}}\cdot
\frac{8x}{\sin 8x} \cdot
\frac{\red 1}{\red{8x}}\cdot
\frac{8x}{\sin 8x} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{4x}}{\blue 1}\cdot
\frac{\blue{4x}}{\blue1}\cdot
\frac{\red 1}{\red{8x}}\cdot
\frac{\red 1}{\red{8x}}\cdot
\frac{\sin 4x}{4x} \cdot
\frac{\sin 4x}{4x} \cdot
\frac{8x}{\sin 8x} \cdot
\frac{8x}{\sin 8x} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{16x^2}}{\red{64x^2}}\cdot
\frac{\sin 4x}{4x} \cdot
\frac{\sin 4x}{4x} \cdot
\frac{8x}{\sin 8x} \cdot
\frac{8x}{\sin 8x} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)\\[6pt]
%
& = \lim_{x\to0}\left(%
\frac{\blue{1}}{\red{4}}\cdot
\frac{\sin 4x}{4x} \cdot
\frac{\sin 4x}{4x} \cdot
\frac{8x}{\sin 8x} \cdot
\frac{8x}{\sin 8x} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)\\[6pt]
%
& = \frac{\blue{1}}{\red{4}}\cdot\lim_{x\to0}\left(%
\frac{\sin 4x}{4x} \cdot
\frac{\sin 4x}{4x} \cdot
\frac{8x}{\sin 8x} \cdot
\frac{8x}{\sin 8x} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)
\end{align*}
$$
Step 7
Evaluate the limit of each factor.
$$
\\
\begin{align*}%
&\quad\frac 1 4\cdot\lim_{x\to0}\left(%
\blue{\frac{\sin 4x}{4x}} \cdot
\blue{\frac{\sin 4x}{4x}} \cdot
\red{\frac{8x}{\sin 8x}} \cdot
\red{\frac{8x}{\sin 8x}} \cdot
\frac{1+\cos 8x}{1+\cos 4x}
\right)\\[6pt]
%
& = \frac 1 4 (\blue 1)(\blue 1)(\red 1)(\red 1)\left(\frac{1 + \cos 0}{1 + \cos 0}\right)\\[6pt]
%
& = \frac 1 4
\end{align*}
\\
$$
Answer
$$
\displaystyle \lim_{x\to0}\,\frac{1-\cos 4x}{1 - \cos 8x} = \frac 1 4
$$
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