limx→1f(x)\mathop {\lim }\limits_{x \to 1} f\left( x \right)x→1limf(x) no existe porque limx→1−f(x)=−2≠limx→1+f(x)=2\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = - 2 \ne \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = 2x→1−limf(x)=−2=x→1+limf(x)=2