If ` f(x)={{:(|x|+1,xlt0),(0,x=0) ,(|x|-1,xgt0):}`
for what value (s) of a does `underset(xrarra)"lim"f(x)` exists?

Explanation

Case 1. When `a lt 0.`
In this case, we have
`underset(xtoa^(+))(lim)f(x)=underset(hto0)(lim)f(a+h)=underset(hto0)(lim){|a+h|+1}=|a|+1.`
`underset(xtoa^(+))(lim)f(x)underset(hto0)(lim)f(a-h)=underset(hto0)(lim){|a-h|+1}=|a|+1.`
`therefore underset(xtoa^(+))limf(x)=underset(xtoa)limf(x)=|a|+1.`
So, in this case, `underset (xtoa)lim` f(x)n exists.
Case 2. When `a lt 0.`
In this case, we have
`underset (xtoa^(+))limf(x)=underset(xto0)lim(a+h)=underset(hto0)lim{|a-h|-1}=|a|-1.`
`underset(xtoa^(+))limf(x)=underset(hto0)limf(a-h)=underset(hto0)lim{|a-h|-1}=|a|-1.`
`thereforeunderset(xtoa^(+))limf(x)=underset(xto0)limf(x)=|a|-1.`
So, in this case, `unederset(xtoa)lim f(x)` exisets.
Case 3.
When `a lt 0.`
In this case, we have
`underset(xto0^(-))limf(x)=underset(hto0)limf(a+h)=underset(hto0)lim{|a-h|-1}=|a|-1.`
`underset(xto0)limf(x)=underset(hto0)limf(0-h)=underset(hto0)lim{|-h|+1}=1.`
Thus, Thus ` underset(xto0^(+))limf(x)neunderset(xto0^(-))limf(x).`
Hence,` underset(xto0)limf(x)` does not exists.
Thus, `underset(xtoa)limf(x)` exists only when `a ne 0.`