Question

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

Answer 1

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\frac{3x}{-x+2x}={lim}_{x\to 0^{-}}\frac{3x}{x}=3$

$[\because asx\to 0^{-},|x|=-x]{lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\frac{3x}{x+2x}=1$

$[\because asx\to 0^{-},|x|=-x]$

thus, ${lim}_{x\to 0^{+}}f\left(x\right)\ne {lim}_{x\to 0^{+}}f\left(x\right)$

$\therefore {lim}_{x\to 0}f\left(x\right)$ does not exist.