f is defined in \[\left[ { - 5,5} \right]\] as
\[f\left( x \right) = x\] if x is rational
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\] =−x if x is irrational. Then
Solutions · 1
Official solution
Let a is a rational number other than 0, in
\[\left[ { - 5,5} \right],\] then
\[f\left( a \right) = a\] and \[\mathop {\lim }\limits_{x \to a} \,\,f\left( x \right) = - a\]
[ As in the immediate neighbourhood of a rational -
number, we find irrational numbers ]
\[\therefore\] f(x) is not continuous at any rational number
If a is irrational number, then
\[\,f\left( a \right) = - a\] and \[\mathop {\lim }\limits_{x \to a} \,f\left( x \right) = a\]
\[\therefore\] f(x) is not continuous at any irrational number clearly
\[\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 0\]
\[\therefore\] f(x) is continuous at x=0