After trying to work through some basic proofs of limits, I realized that I have some deficits in my understanding of the difference between equality and existence.
Consider the following two statements:
$$\displaystyle \lim_{x \to a} f(x) \text { exists}$$
$$\displaystyle \lim_{x \to a} f(x) = L$$
Now, consider the next two sets of statements:
- $$g(x) \text { exists}$$
- $$g(x)=L$$
Next, consider these two statements:
$$ f(x) \neq 0 \land g(x) \text { exists } \iff \frac{f(x)g(x)}{f(x)} \text{ exists}$$
$$ f(x) \neq 0 \land g(x)=L \iff \frac{f(x)g(x)}{f(x)}=L $$
Finally, consider these two statements:
$$ \displaystyle \lim_{x \to a} f(x) \neq 0 \ \land \ \displaystyle \lim_{x \to a} g(x) \text{ exists} \iff \displaystyle \lim_{x \to a} \frac{f(x)g(x)}{f(x)} \text {exists}$$
$$\displaystyle \lim_{x \to a} f(x) \neq 0 \ \land \ \displaystyle \lim_{x \to a} g(x) =L \iff \displaystyle \lim_{x \to a} \frac{f(x)g(x)}{f(x)} =L$$
The reason for the above comments is to better understand the logic that I see employed in proofs that read, for example, like the following question:
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
The straightforward approach for this question works out as:
...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: $$\displaystyle\lim_{x \to 0}g(x)=\frac{1}{\alpha}\cdot\displaystyle\lim_{x \to 0}f(x)g(x)$$ $\color{red}{\text {Thus, if we assert that} \displaystyle\lim_{x \to 0}g(x) \text{ does not exist, then it must be the case that} \displaystyle\lim_{x \to 0}f(x)g(x) \text{ does not exist}}$.
My question is about the final $\color{red}{\text{red}}$ statement. This conclusion only makes sense to me if the structure of the proof concludes that:
$$ \displaystyle \lim_{x \to 0} f(x) \neq 0 \ \land \ \displaystyle \lim_{x \to 0} g(x) \text{ exists} \iff \displaystyle\lim_{x \to 0}f(x)g(x) \text{ exists}$$
The reason the $\color{red}{\text {red}}$ claim makes sense relative to this biconditional is because if the first part of the biconditional is false (by virtue of the assertion that $\displaystyle \lim_{x \to 0} g(x) \text{ does not exist}$ ), then the second part of the biconditional is false...which means that $\displaystyle\lim_{x \to 0}f(x)g(x) \text{ does not exist}$
However, I am having difficulties seeing how exactly the proof arrived at this aforementioned biconditional. To me, it seems that the proof structure concludes with:
$$ \displaystyle \lim_{x \to 0} f(x) \neq 0 \ \land \ \displaystyle \lim_{x \to 0} g(x) = L \iff \frac{1}{\alpha}\cdot\displaystyle\lim_{x \to 0}f(x)g(x)=L$$
If we assert that $\displaystyle \lim_{x \to 0} g(x) \text{ does not exist}$, then certainly the first part of the biconditional is false. But negating the second part of the biconditional simply produces the statement: $\frac{1}{\alpha}\cdot\displaystyle\lim_{x \to 0}f(x)g(x)\neq L$. This is quite different than claiming that $\displaystyle\lim_{x \to 0}f(x)g(x) \text{ does not exist}$.
Could someone please help me understand what exactly is going on in this proof? Thank you!
Edit: An alternative (and to me more clear) way of writing this proof is as follows:
Suppose $\displaystyle \lim_{x \to 0} f(x)g(x)=L$
Next, consider that, by assumption, $\displaystyle \lim_{x \to 0} f(x) = M \neq 0$. Therefore, $\displaystyle \lim_{x \to 0} \frac{1}{f(x)}=\frac{1}{M}$.
It is then the case that $\displaystyle \lim_{x \to 0} \frac{f(x)g(x)}{f(x)}=\frac{L}{M}$.
But $\frac{f(x)g(x)}{f(x)}=g(x)$ so we could equivalently write:
$\displaystyle \lim_{x \to 0}g(x)=\frac{L}{M}$
This is a contradiction to our initial assumption that $\displaystyle \lim_{x \to 0}g(x) \text{ does not exist}$.
