Answer
The value of the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$ is 1.
Work Step by Step
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.
