Proof of $\lim_{x \to c}f(x).g(x) =\infty$ yesterday my prof gave us a quiz, and I stumbled upon this certain question that I'm not confident on my answer

Suppose that $\lim_{x \to c}f(x) = L$, where $L > 0$, and that $\lim_{x \to c}g(x) = \infty$. Show that $\lim_{x \to c}f(x).g(x) =\infty$. If $L = 0$, show by example that this conclusion may fail.

Now, here's the answer I've come up with.
I've recalled the property of $\lim_{x \to c}f(x).g(x) =\lim_{x \to c}f(x).\lim_{x \to c}g(x)$
then by substitution I got
$\lim_{x \to c}f(x).g(x) = L . \infty = \infty$
Is this really it? because I thought that the correct answer was to use the epsilon-delta definition, but I'm not sure on how to work with that.
Also, for the counterexample part, I chose
$f(x) = \frac{1}{x}$ and  $g(x) = x$ , then $f(x).g(x) = 1$, I get
$\lim_{x \to \infty}f(x) = 0$ and $\lim_{x \to \infty}g(x) = \infty$
then $\lim_{x \to \infty} f(x) .g(x)= 1$, proven fail.
Is this correct?
Any hints or tips would help, thanks beforehand.
 A: Your example is correct. For the first part let $M$ be any positive number. There exist $\delta_1 >0$ such that $|x-c| <\delta_1$ implies $|f(x)-L| <\frac L 2$. This implies that $f(x) >\frac L 2$ if $|x-c| <\delta_1$. Also there exist $\delta_2 >0$ such that $|x-c| <\delta_2$ implies $g(x)>\frac {2M} L$. Let $\delta$ be the minimum of $\delta_1$ and $\delta_2$. Then $|x-c| <\delta$ implies $f(x)g(x)>\frac L 2 (\frac {2M} L)=M$. This proves that $f(x)g(x) \to \infty$ as $x \to c$.
A: The property that you used is not valid because it assumes that the two component limits exist. So, you can see that the limit of a product is the product of the limits when the limits exist. Over here, one of the limits tends to $$+\infty$$ so you need to be a bit more careful.
You can do this easily using a formal argument. Let $M \in \mathbb{R}$. Then, we need to show that:
$$\exists \delta \in \mathbb{R}: 0 < |x-c| < \delta \implies f(x)g(x) > M$$
Since $\lim_{x \to c} g(x) = \infty$, it follows that:
$$\exists \delta_1 > 0: 0 < |x-c| < \delta_1 \implies g(x) > \frac{2M}{L}$$
Since $\lim_{x \to c} f(x) = L$, it follows that:
$$\exists \delta_2: 0 < |x-c| < \delta_2 \implies |f(x)-L| < \frac{L}{2}$$
So, that means that:
$$0 < |x-c| < \delta_2 \implies f(x) > \frac{L}{2}$$
Define $\delta = \min \{\delta_1,\delta_2\}$. Then:
$$0 < |x-c| < \delta \implies f(x)g(x) > \frac{2M}{L} \cdot \frac{L}{2} = M$$
which proves that $$\lim_{x \to c} f(x)g(x) = +\infty$$.
A: For an easy counterexample, let
$$f(x):=(x-c)^2$$ and
$$g(x):=\frac a{(x-c)^2}$$ where $a$ is your favorite constant.
We have
$$\lim_{x\to c}f(x)=0$$ and $$\lim_{x\to c}g(x)=\infty.$$
You can conclude.

Short explanation about the main claim:
As $f$ tends to $L$, you will find neighborhoods of $c$ where $f$ is nonzero and keeps the same sign (the sign of $L$). Then as $g(x)$ tends to infinity, multiples of $g(x)$ also tend to infinity.
