Solución

Utilizando el cálculo de límites y el gráfico de referencia, llegamos a los siguientes valores:

Apartado a

lim⁡x→−4−f(x)=0            lim⁡x→−4+f(x)=0            lim⁡x→−4f(x)=0               f(−4)=0\mathop {\lim }\limits_{x \to - {4^ - }} f\left( x \right) = 0\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to - {4^ + }} f\left( x \right) = 0\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to - 4} f\left( x \right) = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f\left( { - 4} \right) = 0

Apartado b

lim⁡x→−2−f(x)=3            lim⁡x→−2+f(x)=3            lim⁡x→−2f(x)=3               f(−2)  no  definido\mathop {\lim }\limits_{x \to - {2^ - }} f\left( x \right) = 3\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to - {2^ + }} f\left( x \right) = 3\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to - 2} f\left( x \right) = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f\left( { - 2} \right)\,\,no\,\,definido

Apartado c

lim⁡x→1−f(x)=6            lim⁡x→1+f(x)=3            lim⁡x→1f(x)  ∄             f(1)  =6\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 6\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = 3\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right)\,\,\nexists\,\,\,\,\,\,\,\,\,\,\,\,\,f\left( 1 \right)\,\, = 6

Apartado d

lim⁡x→3−f(x)=−∞            lim⁡x→3+f(x)=−∞            lim⁡x→3f(x)=−∞               f(1)  no  definido\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = - \infty \,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = - \infty \,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to 3} f\left( x \right) = - \infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f\left( 1 \right)\,\,no\,\,definido