IF `f(x)={{:(x,"if",x,"is rational"),(1-x,"if",x, "is irrational"):},"then find" lim_(xto1//2) f(x)` if exists.
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`underset(xto1//2^(+))limf(x)=underset(xto1//2^(+))limx" "`(if `1//2^(+)`rational) `=1//2` `underset(xto1//2^(-))limf(x)=underset(xto1//2^(+))lim(1-x)" "`(if `1//2^(+)`is irrational) `=1-1//2=1//2` So, `underset(xto1//2^(+))limf(x)=1//2` in any case. Similarly, we get `underset(xto1//2^(-))limf(x)=1//2` `:." "underset(xto1//2)limf(x)=1//2`
