Find $\lim\limits_{x \to 0} \; \dfrac{3x + \left|x\right|}{2x-\left|x\right|}$, if it exists.
When $x>0, \; \left|x\right|=+x$
When $x<0, \; \left|x\right|=-x$
$\begin{aligned}
\therefore \lim\limits_{x \to 0} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|} & = \lim\limits_{x \to 0^+} \; \dfrac{3x+x}{2x-x} \;\; for \; x > 0 \\
& = \lim\limits_{x \to 0^+} \; \dfrac{4x}{x} \\
& = \lim\limits_{x \to 0^+} 4 = 4
\end{aligned}$
$\begin{aligned}
\text{Also, } \lim\limits_{x \to 0} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|} & = \lim\limits_{x \to 0^-} \; \dfrac{3x-x}{2x+x} \;\; for \; x < 0 \\
& = \lim\limits_{x \to 0^-} \; \dfrac{2x}{3x} \\
& = \lim\limits_{x \to 0^-} \dfrac{2}{3} = \dfrac{2}{3}
\end{aligned}$
$\therefore$ $\lim\limits_{x \to 0^+} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|} \neq \lim\limits_{x \to 0^-} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|}$
$\implies$ $\lim\limits_{x \to 0} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|}$ does not exist.