This exercise is a generalization of the product rule of limits and the case a) given here is actually more powerful:
Generalized Product Rule of Limits: Let $\lim_{x\to a} f(x) =L\neq 0$ then the limiting behavior of $f(x) g(x) $ is same as that of that $g(x) $ when $x\to a$. In other words
- If $\lim_{x\to a} g(x) =M$ then $\lim_{x\to a} f(x) g(x) = LM$. This is the usual product rule given in most textbooks.
- If $g(x) $ diverges to $\pm\infty$ as $x\to a$ then so does $f(x) g(x) $ as $x\to a$ (the sign of $\infty$ being same or different according as $L>0$ or $L<0$).
- If $g(x) $ oscillates finitely or infinitely as $x\to a$ then so does $f(x) g(x)$.
Part a) of the question combines the second and third bullet points of the above result into one. The proof of this is done using contradiction and quotient rule. Suppose on the contrary that limit of $fg$ exists then $g=fg/f$ and via quotient rule limit of $g$ also exists. Notice that the limit $L$ of $f$ must be non-zero in order to apply the quotient rule.
In order to prove the above generalized theorem we need to separately prove the second and third bullet point using definitions. This is not really difficult and you should try it.
Part b) of the question says that the second and third bullet points hold even if $|f|\to \infty$. But the more important part is that if $|f|\to\infty$ then the first bullet point above does not hold necessarily. This is also proved in the same manner.
For c) let $f(x) =x, g(x) =1/x$.
The fact that $f$ has a non zero limit is more powerful because it allows us to infer the behavior of product $fg$ on the basis of behavior of $g$ in all cases (whereas $|f|\to\infty$ helps us only in those non-interesting cases of non-existence). And the product rule of limits should be used in this generalized form while evaluating limits in step-by-step fashion.
Thus if we have a complicated expression whose limit we wish to evaluate and we are able to write it in the form $f(x) g(x) $ where $f$ is simple and $g$ is less complicated than original expression then it is really a great help if $f$ tends to a non-zero limit $L$. In that case the problem is reduced to $L\cdot\lim g$ and thus the expression gets less complicated in one step. This step is performed without any idea about limit of $g$ and this is justified only by the above mentioned generalized product rule. A similar rule is available for sum $f+g$ where only the existence of limit of $f$ is needed.
