## Precalculus (6th Edition) Blitzer

Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$. In making the table, choose the value of x close to 0 from the left and from the right as x approaches 0. As x approaches 0 from the left, arbitrarily start with $x=-1$. Then select two additional values of x that are closer to 0, but still less than 0; we choose $-0.1$ and $-0.01$. Evaluate f at each chosen value of x to obtain the corresponding values of $f\left( x \right)$. All the values of x are in radians. At $x=-1$, Substitute, $x=-1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$. Therefore, $\underset{x\to -1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( -1 \right)}{-1}$ Apply the trigonometric property, $\tan \left( -\theta \right)=-\tan \left( \theta \right)$. \begin{align} & \underset{x\to -1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{-\tan \left( 1 \right)}{-1} \\ & =1.5574 \end{align} Now, at $x=-0.1$ Substitute, $x=-0.1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$. $\underset{x\to -0.1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( -0.1 \right)}{-0.1}$ Apply the trigonometric property, $\tan \left( -\theta \right)=-\tan \left( \theta \right)$ \begin{align} & \underset{x\to -0.1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{-\tan \left( 0.1 \right)}{-\left( 0.1 \right)} \\ & =\frac{0.1003}{0.1} \\ & =1.0033 \end{align} At $x=-0.01$ Substitute, $x=-0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$. $\underset{x\to -0.01}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( -0.01 \right)}{-\left( 0.01 \right)}$ Apply the trigonometric property, $\tan \left( -\theta \right)=-\tan \left( \theta \right)$ \begin{align} & \underset{x\to -0.01}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{-\tan \left( 0.01 \right)}{-\left( 0.01 \right)} \\ & =\frac{0.0100}{0.01} \\ & =1.00003 \end{align} Now, as x approaches $0$ from the right, arbitrarily start with $x=1$. Then select two additional values of x that are closer to $0$, but still greater than $0$; we choose $0.1$ and $0.01$. At $x=1$ Substitute, $x=1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$. Therefore, \begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( 1 \right)}{1} \\ & =\frac{1.5574}{1} \\ & =1.5574 \end{align} Now, at $x=0.1$ Substitute, $x=0.1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$. Therefore, \begin{align} & \underset{x\to 0.1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( 0.1 \right)}{0.1} \\ & =\frac{0.1003}{0.1} \\ & =1.0033 \end{align} At $x=0.01$ Substitute, $x=0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$. \begin{align} & \underset{x\to 0.01}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( 0.01 \right)}{0.01} \\ & =\frac{0.0100}{0.01} \\ & =1.00003 \end{align} The limit of $\frac{\tan x}{x}$ as x approaches $0$ equals the number $1$.