Find $\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{6x}$.

Solution We follow the steps for finding a limit using L’Hôpital’s Rule.

Step 1 Since $\lim\limits_{x\rightarrow 0}\tan x=0$ and $\lim\limits_{x\rightarrow 0}( 6x) =0$, the quotient $\dfrac{\tan x }{6x}$ is an indeterminate form at $0$ of the type $\dfrac{0}{0}$.

Step 2 $\dfrac{d}{\textit{dx}}\tan x=\sec ^{2}x$ and $\dfrac{d}{\textit{dx}}\left( 6x\right) =6$.

Step 3 $\lim\limits_{x\rightarrow 0}\dfrac{\dfrac{d}{\textit{dx}}\tan x}{\dfrac{d}{dx}\left( 6x\right) }=$ $\lim\limits_{x\rightarrow 0}\dfrac{\sec ^{2}x}{6}=\dfrac{1}{6}.$

It follows from L’Hôpital’s Rule that $\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{6x}=\dfrac{1}{6}$.