Proof of $\lim_{x \to c}f(x).g(x) =\infty$

Question

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.

Answer 1

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 $$\lim_{x \to c} f(x)g(x) = +\infty$$.

Answer 2

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 $x \to c$.

Answer 3

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.