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Suppose $\lim_{x\to 0} |f(x)|=\infty$ and $\lim_{x\to 0} g(x) $ does not exist. Prove that $\lim_{x\to a} f(x)g(x) $ also does not exist.

Asume $\lim_{x\to a} f(x)g(x) $ exist. I have $|x|< \delta_1$ then $|f(x)|>N$ for every $N > 0$ which implies $|f(x)g(x)|> N|g(x)|$. I also have $|f(x)g(x) - l|< \epsilon$ as $|x|< \delta_2$ for some $\delta_2$. Now if $|x|<\min(\delta_1,\delta_2) \to$ $ N|g(x)|<|f(x)g(x)|<\epsilon +|l|$ which tell use $|g(x)|<\frac {\epsilon +|l|}{N}$ and $\lim_{x\to 0} g(x) = 0 $. Can $|g(x)|<\frac {\epsilon +|l|}{N}$ be accepted that $|g(x)|$ is arbitrarily small?

Arturo Magidin
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Christopher
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1 Answers1

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It seems to me that you intuitively know what you need to prove and how to prove it; you have (at least conceptually) the essential pieces of a proof, you understand fundamentally how they fit together, and the question is just a detail on how to write it up.

If you must do a new epsilon-delta proof for a theorem like this rather than using already proved facts, I think the usual practice is to subscript "epsilons" as well as "deltas" when applying the given limits. For example, you could say that for any positive $\epsilon_2,$ there exists a $\delta_2$ such that $\lvert f(x)g(x)\rvert < \epsilon_2$ whenever $\lvert x\rvert < \delta_2$.

Then when $\lvert x\rvert < \delta = \min\{\delta_2,\delta_2\}$, you have shown that $\lvert g(x)\rvert < \dfrac{\epsilon_2 + \lvert l\rvert}{N}.$

The only thing remaining to wrap this up in the usual form of an epsilon-delta proof is to explain how, for any given $\epsilon > 0,$ you would choose $\delta_1$ and $\delta_2$ so as to make $\dfrac{\epsilon_2 + \lvert l\rvert}{N}$ less than or equal to $\epsilon$. And in order to do this, the statements about the limits of $f(x)$ and $f(x)g(x)$ allow you to set $\epsilon_2$ and $N$ to any positive values you want.

I don't think it invalidates your proof if you use symbols and language in a way differently from the usual form, but it is easier for people to understand your proof and verify it if you use the usual form.

David K
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