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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?

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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 $\lim_{x\to0}|f(x)|=\infty$, $\lim_{x\to0}g(x)=0$.

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  • $\begingroup$ 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" $\endgroup$ – S.Cramer Apr 20 at 8:49
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos Apr 20 at 8:50
  • $\begingroup$ 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? $\endgroup$ – S.Cramer Apr 20 at 9:03
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    $\begingroup$ Yes, that is correct. $\endgroup$ – José Carlos Santos Apr 20 at 9:34
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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 $\lim\limits_{x \to 0}\frac{1}{|f(x)|}=0$.

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