Find \(\lim\limits_{x\rightarrow 4}\dfrac{\sqrt{3x^{2}+1}}{x-1}\).

Solution We seek the limit of the quotient of two functions. Since the limit of the denominator \(\lim\limits_{x\rightarrow 4}(x-1)\neq 0\) , we use the Limit of a Quotient. \[ \begin{eqnarray*} \lim_{x\rightarrow 4}\frac{\sqrt{3x^{2}+1}}{x-1}\underset{\underset{\color{#0066A7}{\hbox{Limit of a Quotient}}}{\color{#0066A7}{\uparrow}}}{=}\frac{\lim\limits_{x\rightarrow 4}\sqrt{ 3x^{2}+1}}{\lim\limits_{x\rightarrow 4}(x-1)}\underset{\underset{\color{#0066A7}{\hbox{Limit of a Root}}}{\color{#0066A7}{\uparrow}}}{=}\frac{\sqrt{\lim\limits_{x\rightarrow 4} (3x^{2}+1)}}{\lim\limits_{x\rightarrow 4}(x-1)}=\frac{\sqrt{3\cdot 4^{2}+1}}{4-1}=\frac{\sqrt{49}}{3}=\frac{7}{3}\\[-13pt] \end{eqnarray*}\]