# Manipulating limit expressions when you are told that the limit of one of the expressions does not exit (Spivak - Chapter 5 Problem 23)

Questions 23 a) and b) from Chapter 5 of of Spivak's Calculus are written as follows:

a) Suppose that $$\displaystyle\lim_{x \to 0}f(x)$$ exists and is $$\neq 0$$. Prove that if $$\displaystyle\lim_{x \to 0}g(x)$$ does not exist, then $$\displaystyle\lim_{x \to 0}f(x)g(x)$$ does not exist.

b) Prove the same result if $$\displaystyle\lim_{x \to 0}\ \lvert f(x)\rvert=\infty$$.

The method for a) is to recognize that, because $$\displaystyle\lim_{x \to 0}f(x)\neq0$$, the following expression simplifies accordingly:

$$\displaystyle\lim_{x \to 0}g(x)=\displaystyle\lim_{x \to 0}\frac{f(x)g(x)}{f(x)}=\displaystyle\lim_{x \to 0}f(x)g(x)\cdot\frac{1}{\displaystyle\lim_{x \to 0}f(x)}$$

Letting $${\displaystyle\lim_{x \to 0}f(x)}=\alpha$$, we then get:

$$\displaystyle\lim_{x \to 0}g(x)=\frac{1}{\alpha}\cdot\displaystyle\lim_{x \to 0}f(x)g(x)$$

Thus, if we assert that $$\displaystyle\lim_{x \to 0}g(x)\ \text{DNE}$$, then it must be the case that $$\displaystyle\lim_{x \to 0}f(x)g(x)\ \text{DNE}$$.

This proof strikes me as a little weird because the manipulations I carried out (when rearranging the terms) assumed that the limits existed. Is this acceptable?

(i.e. Does it even make sense to manipulate limit expressions if I know that the expression in question does not have a limit?)

Now, when it comes to b), I am not really certain of how to structure my proof. My understanding of the phrase:

$$\displaystyle\lim_{x \to 0}\ (\cdot)=\infty$$

is that the limit of $$(\cdot) \ \text{DNE}$$. I recognize that there is a more technical definition, but I believe the statement of "$$\text {DNE}$$" is nonetheless valid.

As such, in the case of $$\displaystyle\lim_{x \to 0}\ \lvert f(x)\rvert=\infty$$, I am reluctant to recreate my former argument by bringing $${\displaystyle\lim_{x \to 0}f(x)}$$ into the denominator. Any suggestions?

Your proof is fine. It proves (correctly) that, if the limits $$\lim_{x\to0}f(x)g(x)$$ and $$\lim_{x\to0}f(x)$$ exist, then the limit $$\lim_{x\to0}g(x)$$ exists too. But you are assuming that it doesn't.
Concerning b), if $$\lim_{x\to0}f(x)g(x)=l\in\Bbb R$$, then, since $$\lim_{x\to0}|f(x)|=\infty$$, $$\lim_{x\to0}g(x)=0$$.
• For your solution to b), if I assume that $\displaystyle\lim_{x \to 0} g(x)$ does not exist, then is it also the case that $\displaystyle\lim_{x \to 0} f(x)g(x)$ does not exist? Your version is the contrapositive, yes? If so, your version is really saying "if $\displaystyle\lim_{x \to 0} f(x)g(x)$ exists, then $\displaystyle\lim_{x \to 0} g(x)$ exists" – S.Cramer Apr 20 at 8:49
• Yes: if $\lim_{x\to0}g(x)$ does not exist, then $\lim_{x\to0}f(x)g(x)$ does not exist (in $\Bbb R$). And, yes, my version is the contrapositive. – José Carlos Santos Apr 20 at 8:50
• Final clarification (if you're willing). When I originally compare $\displaystyle\lim_{x \to 0}g(x)$ and $\displaystyle\lim_{x \to 0}\frac{f(x)g(x)}{f(x)}$, I am effectively saying $\displaystyle\lim_{x \to 0}g(x) \text { exists} \iff \displaystyle\lim_{x \to 0}\frac{f(x)g(x)}{f(x)} \text { exists}$, correct? This is why I can follow my conclusion to a) with, "Thus, if $\displaystyle\lim_{x \to 0}g(x) \text { exists}$ is false, ..." etc. Is that correct? – S.Cramer Apr 20 at 9:03
For b): $$\lim\limits_{x \to 0}\ \lvert f(x)\rvert=\infty$$ gives, that in some neighbourhood of $$0$$ we have $$f(x)\ne 0$$ so we have right to consider $$\frac{1}{f(x)}$$ there. So here will work same contradiction logic as in a). By the way $$\lim\limits_{x \to 0}\frac{1}{|f(x)|}=0$$.