Solución

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

Apartado a

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

Apartado b

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

Apartado c

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

Apartado d

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