Answer
(a) $\lim\limits_{x \to a} [f(x) + g(x)] = \infty$
(b) $\lim\limits_{x \to a} [f(x)g(x)] = \infty~~$ if $c \gt 0$
(c) $\lim\limits_{x \to a} [f(x)g(x)] = -\infty~~$ if $c \lt 0$
Work Step by Step
(a) Let $\epsilon = 1$
There is a number $\delta_1$ such that if $\vert x - a \vert \lt \delta_1$, then $\vert g(x)-c \vert \lt 1$
Then $~~-\vert c \vert -1 \lt g(x) \lt \vert c \vert + 1$
Choose any positive number $M$
There is a number $\delta_2$ such that if $\vert x - a \vert \lt \delta_2$, then $f(x) \gt M+\vert c \vert + 1$
Let $\delta = min\{\delta_1, \delta_2\}$
Then:
If $\vert x-a \vert \lt \delta$, then $f(x)+g(x) \gt (M+\vert c\vert + 1) -\vert c \vert - 1 = M$
Therefore, $\lim\limits_{x \to a} [f(x) + g(x)] = \infty$
(b) Suppose that $c \gt 0$
Let $\epsilon = \frac{c}{2}$
There is a number $\delta_1$ such that if $\vert x - a \vert \lt \delta_1$, then $\vert g(x)-c \vert \lt \frac{c}{2}$
Then $~~\frac{c}{2} \lt g(x) \lt \frac{3c}{2}$
Choose any positive number $M$
There is a number $\delta_2$ such that if $\vert x - a \vert \lt \delta_2$, then $f(x) \gt \frac{2M}{c}$
Let $\delta = min\{\delta_1, \delta_2\}$
Then:
If $\vert x-a \vert \lt \delta$, then $f(x)g(x) \gt \frac{2M}{c}\cdot \frac{c}{2} = M$
Therefore, $\lim\limits_{x \to a} [f(x)g(x)] = \infty$
(c) Suppose that $c \lt 0$
Let $\epsilon = \vert \frac{c}{2} \vert$
There is a number $\delta_1$ such that if $\vert x - a \vert \lt \delta_1$, then $\vert g(x)-c \vert \lt \frac{c}{2}$
Then $~~\frac{3c}{2} \lt g(x) \lt \frac{c}{2}$
Choose any negative number $N$
There is a number $\delta_2$ such that if $\vert x - a \vert \lt \delta_2$, then $f(x) \gt \frac{2 N }{ c}$
Note that $\frac{2N}{c}$ is a positive number since $N$ and $c$ are both negative numbers.
Let $\delta = min\{\delta_1, \delta_2\}$
Then:
If $\vert x-a \vert \lt \delta$, then $f(x)g(x) \lt \frac{2N}{c}\cdot \frac{c}{2} = N$
Therefore, $\lim\limits_{x \to a} [f(x)g(x)] = -\infty~~$ if $c \lt 0$
