Question

If $f\left(x\right)=$ $\{\begin{array}{cc}\left|x\right|+1,& x<0\\ 0,& x=0\\ x^3\left|x\right|-1& x>0\end{array}$
For what value (s) of a does $\underset{x\to a}{lim}f\left(x\right)$ exists?

Answer 1

The given function is
$f\left(x\right)=$ $\{\begin{array}{cc}\left|x\right|+1,& x<0\\ 0,& x=0\\ x^3\left|x\right|-1& x>0\end{array}$
When $a=0$
$\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{-}}{lim}(\left|x\right|+1)$
= $\underset{x\to 0}{lim}(-x+1)$ $[Ifx<0,\left|x\right|=-x]$
$=-0+1$
$=1$
$\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}(\left|x\right|-1)$
= $\underset{x\to 0}{lim}(x-1)$ $[Ifx<0,\left|x\right|=x]$
$=0-1$
$=-1$
Here it is observed that $\underset{x\to a}{lim}f\left(x\right)\ne \underset{x\to a^{+}}{lim}f\left(x\right)$
$\therefore \underset{x\to 0}{lim}f\left(x\right)$ does not exist
When $a<0$
$\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{-}}{lim}(\left|x\right|+1)$
= $\underset{x\to a}{lim}(-x+1)$ $[x<a<0\Rightarrow \left|x\right|=-x]$
$=-a+1$
$\underset{x\to a^{+}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}(\left|x\right|+1)$
$\underset{x\to a}{lim}(-x+1)$ $[a<x<0\Rightarrow \left|x\right|=-x]$
$=-a+1$
$\therefore \underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=-a+1$

Thus limit of $f\left(x\right)$ exists at $x=a$ where $a<0$
When $a>0$
$\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{-}}{lim}(\left|x\right|-1)$
= $\underset{x\to a}{lim}(x-1)$ $[0<x<a\Rightarrow \left|x\right|=x]$
$=a-1$
$\underset{x\to a^{+}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}(\left|x\right|-1)$
= $\underset{x\to a}{lim}(x-1)$ $[0<a<x\Rightarrow \left|x\right|=x]$
$=a-1$
$\therefore \underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=a-1$
Thus limit of $f\left(x\right)$ exists at $x=a$ where $a>0$
Thus $\underset{x\to }{lim}af\left(x\right)$ exists for all $a\ne 0\cdot$