## Precalculus (6th Edition) Blitzer

The value of the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$ is 1.
Consider the limit to be solved, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$. Substitute $\tan x=\frac{\sin x}{\cos x}$ in $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$: \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{\sin x}{\cos x}}{x} \\ & =\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x\times \cos x} \\ & =\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}\times \frac{1}{\cos x} \end{align} Using $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)\times g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\times \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}\times \frac{1}{\cos x} \\ & =\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\cos x} \end{align} Using quotient property of limits $\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\frac{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}{\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)},\ \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)\ne 0$: \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\cos x} \\ & =\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}\times \frac{\underset{x\to 0}{\mathop{\lim }}\,1}{\underset{x\to 0}{\mathop{\lim }}\,\cos x} \end{align} Use the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1\ \text{and }\underset{x\to 0}{\mathop{\lim }}\,\cos x=1$. Also, use limit property $\underset{x\to a}{\mathop{\lim }}\,c=c,c\text{ is constant}$ for $c=1$: \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}\times \frac{\underset{x\to 0}{\mathop{\lim }}\,1}{\underset{x\to 0}{\mathop{\lim }}\,\cos x} \\ & =1\times \frac{1}{1} \\ & =1 \end{align} Thus, the value of the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$ is 1.