Quick Overview
- $$\displaystyle \lim_{\theta\to0} \frac{\sin(\theta)} \theta = 1$$
- This limit was derived in the lesson on the Squeeze Theorem
- The denominator must be the same as the argument of the sine, and both must approach zero in the limit.
Examples
Example 1
Evaluate $$\displaystyle \lim_{\theta\to0}\frac{\sin(4\theta)} \theta$$
Multiply by $$\frac 4 4$$ so the denominator matches the argument.
$$ \begin{align*} \lim_{\theta\to0}\,\frac{\sin(4\theta)} \theta % & = \lim_{\theta\to0} \left(% \frac{\red 4}{\red 4} \cdot \frac{\sin(4\theta)} \theta \right)\\[6pt] % & = \lim_{\theta\to0}\left(% \frac{\red 4} 1\cdot \frac{\sin(4\theta)}{\red 4\theta} \right)\\[6pt] % & = \red{4}\,\displaystyle\lim_{\theta\to0}\, \frac{\sin(4\theta)}{\red 4\theta} \end{align*} $$
Evaluate the limit.
Since the denominator is the same as the argument of the sine function, and both are going to 0, the limit is equal to 1.
$$ 4\,\blue{\displaystyle\lim_{\theta\to0}\, \frac {\sin(4\theta)} {4\theta}} = 4(\blue 1) = 4 $$
Answer: $$\displaystyle \lim_{\theta\to0}\,\frac{\sin(4\theta)} \theta = 4$$
Example 2
Evaluate $$\displaystyle \lim_{x\to0}\,\frac{\sin(5x)}{2x}$$
Factor out the 2 from the denominator.
$$ \displaystyle\lim_{x\to0}\,\frac{\sin(5x)}{\blue{2}x} % = \displaystyle\lim_{x\to0}\left(% \frac 1 {\blue{2}}\cdot \frac{\sin(5x)}{x} \right) % = \frac 1 {\blue 2}\cdot \displaystyle\lim_{x\to0}\,\frac{\sin(5x)} x $$
Multiply by $$\frac 5 5$$ so the denominator matches the argument of the sine function.
\begin{align*} \frac 1 2\cdot\lim_{x\to0}\,\frac{\sin(5x)}{x} % & = \frac 1 2\cdot\lim_{x\to0}\left(% \frac{\red 5}{\red 5}\cdot \frac{\sin(5x)}{x} \right)\\[6pt] % & = \frac 1 2\cdot\displaystyle\lim_{x\to0}\left(% \frac{\red 5} 1\cdot \frac{\sin(5x)}{\red 5 x} \right)\\[6pt] % & =\frac{\red 5} 2 \cdot \lim_{x\to0}\,\frac{\sin(5x)}{\red 5 x} \end{align*}
Evaluate the limit.
$$ \frac 5 2\cdot \blue{\displaystyle\lim_{x\to0}\, \frac{\sin(5x)}{5x}} = \frac 5 2 (\blue 1) = \frac 5 2 $$
Answer: $$\displaystyle \displaystyle\lim_{x\to0}\,\frac{\sin(5x)}{2x} = \frac 5 2$$
Example 3
Evaluate $$\displaystyle \lim_{x\to0} \,\frac{6x}{\sin 3x}$$
Factor a 2 out of the numerator.
$$ \displaystyle\lim_{x\to0} \,\frac{\blue 6 x}{\sin 3x} % = \displaystyle\lim_{x\to0} \left(% \frac{\blue 2} 1 \cdot \frac{\blue 3 x}{\sin 3x} \right) % = \blue 2\cdot \displaystyle\lim_{x\to0}\,\frac{\blue 3 x}{\sin 3x} $$
Rewrite the fraction as its reciprocal to the -1 power.
$$ 2\cdot \displaystyle\lim_{x\to0}\,\frac{3x}{\sin 3x} % = 2\cdot \displaystyle\lim_{x\to0}\,\left(\frac{\sin 3x}{3x}\right)^{-1} $$
Pass the limit inside the exponent (see the page on Limit Laws), and evaluate.
$$ 2\cdot \displaystyle\lim_{x\to0}\,\left(\frac{\sin 3x}{3x}\right)^{-1} % = 2\,\left(\blue{\displaystyle\lim_{x\to0}\frac{\sin 3x}{3x}}\right)^{-1} % = 2(\blue 1)^{-1} % = 2 $$
Answer: $$\displaystyle\, \lim_{x\to0} \,\frac{6x}{\sin 3x} = 2$$
Important! Example 3 shows us that $$\displaystyle \lim_{x\to 0}\, \frac{\theta}{\sin \theta} = 1.$$
Example 4
Evaluate $$\displaystyle \lim_{t\to0} \,\frac{\sin 3t}{\sin 8t}$$
Rewrite as separate fractions.
$$ \displaystyle\lim_{t\to0}\, \frac {\sin 3t} {\sin 8t} = \displaystyle\lim_{t\to0}\, \left(% \frac{\sin 3t } 1 \cdot \frac 1 {\sin 8t } \right) $$
Multiply by $$\frac{3t}{3t}$$ and $$\frac{8t}{8t}$$.
$$ \begin{align*} \lim_{t\to0}\, \left(% \frac{\sin 3t } 1 \cdot \frac 1 {\sin 8t } \right) % & = \lim_{t\to0}\, \left(% \frac{\blue{3t}}{\blue{3t}} \cdot \frac{\sin 3t } 1 \cdot \frac{\red{8t}}{\red{8t}} \cdot \frac 1 {\sin 8t } \right)\\[6pt] % & = \lim_{t\to0}\, \left(% \frac{\blue{3t}}{1} \cdot \frac{\sin 3t }{\blue{3t}} \cdot \frac{1}{\red{8t}} \cdot \frac{\red {8t}} {\sin 8t } \right) \end{align*} $$
Simplify the non-trigonometric fractions
$$ \begin{align*} \lim_{t\to0} \left(% \blue{\frac{3t}{1}} \cdot \frac{\sin 3t}{3t}\cdot \red{\frac{1}{8t}} \cdot \frac{8t} {\sin 8t} \right) % & = \lim_{t\to0} \left(% \blue{\frac{3t}{1}} \cdot \red{\frac{1}{8t}} \cdot \frac{\sin 3t}{3t}\cdot \frac{8t} {\sin 8t} \right)\\[6pt] % & = \lim_{t\to0} \left(% \frac{\blue{3t}}{\red{8t}} \cdot \frac{\sin 3t}{3t}\cdot \frac{8t} {\sin 8t} \right)\\[6pt] % & = \lim_{t\to0} \left(% \frac{\blue{3}}{\red{8}} \cdot \frac{\sin 3t}{3t}\cdot \frac{8t} {\sin 8t} \right)\\[6pt] % & = \frac{\blue{3}}{\red{8}} \cdot \lim_{t\to0} \left(% \frac{\sin 3t}{3t}\cdot \frac{8t} {\sin 8t} \right) \end{align*} $$
Evaluate each of the limits.
$$ \frac 3 8 \left(% \blue{\displaystyle\lim_{t\to0} \frac{\sin 3t}{3t}}\right) \left(\red{\displaystyle\lim_{t\to0} \frac{8t} {\sin 8t}} \right) % = \frac 3 8 (\blue 1)(\red 1) % = \frac 3 8 $$
Answer: $$\displaystyle \lim_{t\to0} \frac{\sin 3t}{\sin 8t} = \frac 3 8$$
Practice Problems
Multiply by $$\frac 7 7$$.
$$ \begin{align*} \lim_{x\to0} \frac{\sin 7x}x % & = \lim_{x\to0} \left(% \blue{\frac 7 7} \cdot \frac{\sin 7x} x \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue 7} 1 \cdot \frac{\sin 7x}{\blue 7 x} \right)\\[6pt] % & = \blue 7\cdot\lim_{x\to0}\,\frac{\sin 7x}{\blue 7 x} \end{align*} $$
Evaluate the limit.
$$7\cdot\blue{\displaystyle\lim_{x\to0} \frac{\sin 7x}{7 x}} = 7(\blue 1) = 7$$
Answer: $$\displaystyle \lim_{x\to0}\, \frac{\sin 7x}x =7$$
Factor a 2 out of the numerator.
$$ \displaystyle\lim_{x\to0} \,\frac{\blue{18}x}{\sin 9x} % = \displaystyle\lim_{x\to0}\left(% \blue{2}\cdot \frac{\blue{9}x}{\sin 9x} \right) % = \blue 2\cdot\displaystyle\lim_{x\to0}\, \frac{\blue{9}x}{\sin 9x} $$
Evaluate the limit.
$$2\cdot\blue{\displaystyle\lim_{x\to0}\, \frac{9x}{\sin 9x}} = 2(\blue 1) = 2$$
Answer: $$\displaystyle \lim_{x\to0}\, \frac{18x}{\sin 9x} = 2$$
Factor the 8 out of the denominator.
$$ \displaystyle\lim_{x\to0}\, \frac{\sin 6x}{\blue 8 x} % = \displaystyle\lim_{x\to0}\left(% \frac 1 {\blue 8} \cdot \frac{\sin 6x} x \right) % = \frac 1 {\blue 8}\cdot\displaystyle\lim_{x\to0}\, \frac{\sin 6x} x $$
Multiply by $$\frac 6 6$$.
$$ \begin{align*} \frac 1 8\cdot\lim_{x\to0}\, \frac{\sin 6x} x % & = \frac 1 8\cdot\lim_{x\to0}\left(% \blue{\frac 6 6}\cdot \frac{\sin 6x} x \right) \\[6pt] % & = \frac 1 8\cdot\lim_{x\to0}\left(% \frac{\blue 6} 1\cdot \frac{\sin 6x}{\blue 6 x} \right)\\[6pt] % & = \frac{\blue 6} 8\cdot\lim_{x\to0}\, \frac{\sin 6x}{\blue 6 x}\\[6pt] % & = \frac 3 4\cdot\lim_{x\to0}\, \frac{\sin 6x}{6 x} \end{align*} $$
Evaluate the limit.
$$ \frac 3 4 \cdot\blue{\displaystyle\lim_{x\to0}\, \frac{\sin 6x}{6x}} % = \frac 3 4(\blue 1) % = \frac 3 4 $$
Answer: $$\displaystyle \lim_{x\to0} \,\frac{\sin 6x}{8x} = \frac 3 4$$
Factor the 5 out of the numerator.
$$ \displaystyle\lim_{x\to0}\, \frac{\blue 5 x}{\sin 9x} % = \displaystyle\lim_{x\to0} \left(% \frac{\blue 5} 1 \cdot \frac x {\sin 9x} \right) % =\blue5\cdot\displaystyle\lim_{x\to0} \,\frac{x}{\sin 9x} $$
Multiply by $$\frac 9 9$$.
$$ \begin{align*} 5\cdot\lim_{x\to0}\,\frac x {\sin 9x} % & = 5\cdot\lim_{x\to0}\left(% \blue{\frac 9 9} \cdot \frac x {\sin 9x} \right)\\[6pt] % & = 5\cdot\lim_{x\to0}\left(% \frac 1 {\blue 9}\cdot \frac{\blue 9 x}{\sin 9x} \right)\\[6pt] % & = \frac 5 {\blue 9}\cdot\lim_{x\to0}\,\frac{\blue 9 x}{\sin 9x} \end{align*} $$
Evaluate the limit.
$$ \frac 5 9\cdot\blue{\displaystyle\lim_{x\to0}\,\frac{9 x}{\sin 9x}} % = \frac 5 9 (\blue 1) % = \frac 5 9 $$
Answer: $$\displaystyle \lim_{x\to0}\, \frac{5x}{\sin 9x} = \frac 5 9$$
Rewrite in separate fractions.
$$ \displaystyle\lim_{x\to0}\,\frac{\sin 4x}{\sin 11x} % = \displaystyle\lim_{x\to0} \left(% \frac{\sin 4x} 1 \cdot \frac 1 {\sin 11x} \right) $$
Multiply by $$\frac{4x}{4x}$$ and $$\frac{11x}{11x}$$.
$$ \begin{align*} \lim_{x\to0} \left(% \frac{\sin 4x} 1 \cdot \frac 1 {\sin 11x} \right) % & = \lim_{x\to0} \left(% \blue{\frac{4x}{4x}}\cdot \frac{\sin 4x} 1 \cdot \red{\frac{11x}{11x}}\cdot \frac 1 {\sin 11x} \right)\\[6pt] % & = \lim_{x\to0} \left(% \frac{\blue{4x}}{1}\cdot \frac{\sin 4x}{\blue{4x}} \cdot \frac 1 {\red{11x}}\cdot \frac{\red{11x}}{\sin 11x} \right)\\[6pt] \end{align*} $$
Simplify the non-trigonometric fractions.
$$ \begin{align*} \lim_{x\to0} \left(% \blue{\frac{4x} 1}\cdot \frac{\sin 4x}{4x} \cdot \red{\frac 1 {11x}}\cdot \frac{11x}{\sin 11x} \right) % & =\lim_{x\to0} \left(% \frac{\blue{4x}}{\red{11x}}\cdot \frac{\sin 4x}{4x}\cdot \frac{11x}{\sin 11x} \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{4}}{\red{11}}\cdot \frac{\sin 4x}{4x} \cdot \frac{11x}{\sin 11x} \right)\\[6pt] % & = \frac{\blue 4}{\red{11}}\cdot \lim_{x\to0}\left(% \frac{\sin 4x}{4x}\cdot \frac{11x}{\sin 11x} \right) \end{align*} $$
Find the limit of each factor.
$$ \frac 4 {11}\cdot\displaystyle\lim_{x\to0}\left(% \blue{\frac{\sin 4x}{4x}}\cdot \red{\frac{11x}{\sin 11x}} \right) % = \left(% \blue{\displaystyle\lim_{x\to0}\,\frac{\sin 4x}{4x}} \right) % \left(% \red{\displaystyle\lim_{x\to0}\,\frac{11x}{\sin 11x}} \right) % = \frac 4 {11}(\blue 1)(\red 1) = \frac 4 {11} $$
Answer: $$\displaystyle \lim_{x\to0}\, \frac{\sin 4x}{11x} = \frac 4 {11}$$
Write the numerator and denominator in separate fractions.
$$ \displaystyle\lim_{x\to0}\,\frac{\sin 5x}{\sin 2x} % = \displaystyle\lim_{x\to0}\left(% \frac{\sin 5x} 1 \cdot \frac 1 {\sin 2x} \right) $$
Multiply by $$\frac{5x}{5x}$$ and $$\frac{2x}{2x}$$.
$$ \begin{align*} \lim_{x\to0}\left(% \frac{\sin 5x} 1 \cdot \frac 1 {\sin 2x} \right) % & = \lim_{x\to0}\left(% \blue{\frac{5x}{5x}}\cdot \frac{\sin 5x} 1 \cdot \red{\frac{2x}{2x}}\cdot \frac 1 {\sin 2x} \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{5x}} 1\cdot \frac{\sin 5x}{\blue{5x}} \cdot \frac 1 {\red{2x}}\cdot \frac{\red{2x}}{\sin 2x} \right) \end{align*} $$
Simplify the non-trigonometric fractions.
$$ \begin{align*} \lim_{x\to0}\left(% \blue{\frac{5x} 1}\cdot \frac{\sin 5x}{5x} \cdot \red{\frac 1 {2x}}\cdot \frac{2x}{\sin 2x} \right) % & = \lim_{x\to0}\left(% \frac{\blue{5x}}{\red{2x}}\cdot \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{5}}{\red{2}}\cdot \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \right)\\[6pt] % & = \frac{\blue{5}}{\red{2}}\cdot\lim_{x\to0}\left(% \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \right) \end{align*} $$
Evaluate the limits of each factor.
$$ \frac 5 2 \cdot\displaystyle\lim_{x\to0}\left(% \blue{\frac{\sin 5x}{5x}} \cdot \red{\frac{2x}{\sin 2x}} \right) % =\frac 5 2\cdot\left(% \blue{\displaystyle\lim_{x\to0}\frac{\sin 5x}{5x}} \right) % \left(% \red{\displaystyle\lim_{x\to0}\frac{2x}{\sin 2x}} \right) % = \frac 5 2 (\blue 1)(\red 1) % = \frac 5 2 $$
Answer: $$\displaystyle \lim_{x\to0}\, \frac{\sin 5x}{\sin 2x} = \frac 5 2$$
