If `f(x)={(x-|x|)/x ,x!=0 ,x=0,s howt h a t("lim")_(xto0)` `f(x)` does not exist.

Explanation

We have
`f(x)={{:(x+1,", "if xge0),(x-1,", "if xlt0):}".`
This function is piecewise function which changes its definition at x=0.
So, to find limiting value of f(x) at x=0 we find LHL and RHL. To the left of zero, we have f(x) `=(x-1)` and to the right of zero, we have f(x) `=(x+1).`
`LHL=underset(xto0)limf(x)=underset(hto0)limf(0-h)=underset(hto0)limf(-h)`
`underset(hto0)lim((-h)-1)`
-1
`RHL=underset(xto0^(+))limf(x)=underset(hto0)limf(0+h)=underset(hto0)limf(h)`
`underset(hto0)lim(h+1)`
`=1`
Thus, LHL`ne`RHL.
Therefore, `underset(xto0)limf(x)` does not exist.