Question

Let $f\left(x\right)=\{\begin{array}{c}x+1,\text{if}x>0\\ x-1,\text{if}x<0\end{array}$ Prove that $li{m}_{x\to 0}$ f(x)does not exist.

Answer 1

$li{m}_{x\to 0^{+}}f\left(x\right)=li{m}_{x\to 0^{+}}(x+1)$

[x=0+h as $x\to 0^{+},h\to 0^{+}]$

$=li{m}_{h\to 0^{+}}(0+h)+1=1$

Also, $li{m}_{x\to 0^{-}}f\left(x\right)=li{m}_{x\to 0^{-}}(x-1)$

[x=0-h as $x\to 0^{-},h\to 0^{+}]$

$li{m}_{x\to 0}(0-h)-1=-1$

$\Rightarrow li{m}_{x\to 0^{+}}f\left(x\right)\ne li{m}_{x\to 0^{-}}f\left(x\right)$

Hence, limit does not exist.