Question

Let f(x) be a function defined by $f\left(x\right)=\left\{\begin{array}{ll}\frac{3x}{\left|x\right|+2x},& x\ne 0\\ 0,& x=0\end{array}\right..$
Show that $\underset{x\to 0}{\lim }f\left(x\right)$ does not exist.

Answer 1

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}\frac{3x}{\left|x\right|+2x},& x\ne 0\\ 0,& x=0\end{array}\right. \\ \mathrm{Left}\mathrm{hand}\mathrm{limit}: \\ \underset{x\to 0^{-}}{\lim }\left[\frac{3x}{\left|x\right|+2x}\right] \\ \mathrm{Let}x=0-h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{3\left(-h\right)}{\left|-h\right|+2\left(-h\right)}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{-3h}{h-2h}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{-3h}{-h}\right] \\ =3 \\ \mathrm{Right}\mathrm{hand}\mathrm{limit}: \\ \underset{x\to 0^{+}}{\lim }\left(\frac{3x}{\left|x\right|+2x}\right) \\ \mathrm{Let}x=0+h,\mathrm{where}h\to 0. \\ {\Rightarrow }^{}\underset{h\to 0}{\lim }\left(\frac{3h}{\left|h\right|+2h}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{3h}{h+2h}\right) \\ =1 \\ \underset{x\to 0^{-}}{\lim }\left(\frac{3x}{\left|x\right|+2x}\right)\ne \underset{x\to 0^{+}}{\lim }\left(\frac{3x}{\left|x\right|+2x}\right) \\ \mathrm{Thus},\underset{x\to 0}{\lim }f\left(x\right)\mathrm{does}\mathrm{not}\mathrm{exist}. \end{gathered}$$