Proof of a Property of Vertical Asymptotes

Question

I'm trying to understand a proof in my Calculus textbook of the following theorem:

Let the functions $f$ and $g$ be continuous on an interval containing $c$. If $f(c) \neq 0$, $g(c) = 0$, and there is an open interval containing $c$ such that $g(x) \neq 0$ for all $x \neq c$ in the interval, then the graph of the function $h(x) = \frac{f(x)}{g(x)}$ has a vertical asymptote at $x = c$.

The proof the textbook provides is the following:

Consider the case for which $f(c) > 0$, and there exists $b > c :$ $c < x < b \implies g(x) > 0$. For $M > 0$, choose $\delta_{1}$ and $\delta_{2}$ such that:

$0<x-c<\delta_1 \implies \frac{f(c)}{2} <f(x)<\frac{3f(c)}{2}$

$0<x-c<\delta_2 \implies 0 < g(x) < \frac{f(c)}{2M}$

Set $\delta = min\{\delta_1, \delta_2\}$. Then $0 < x-c < \delta \implies \frac{f(x)}{g(x)} > \frac{f(c)}{2} \cdot \frac{2M}{f(c)}= > M$, which implies that $\lim_{x \to c^+}\frac{f(x)}{g(x)} = \infty$, so $x = c$ is a vertical asymptote of the graph of $h(x)$.

What I'm wondering is what the point is of

1. Saying that there exists $b > c : c < x < b \implies g(x) > 0$
2. Restricting $f(x) < \frac{3f(x)}{2}$

The way I understand the proof, $\frac{f(c)}{2}$ and $\frac{f(c)}{2M}$ were simply chosen because they cancel nicely when multiplied, correct? But what advantage comes from the other two restrictions I listed?

Any help or explanation would be appreciated, thanks.

Answer 1

There is a significant typo in the greyed region, second displayed line. We want $0\lt g(x)\lt \frac{f(c)}{2M}$. But it looks as if that is not the reason for the concern about the proof.

By assumption there is an interval about $c$ for which $g(x)\ne 0$ except at $c$. The reason for that is that if as $x$ approached $c$, say from the right, $g(x)$ changed sign infinitely often (which can happen), then $\frac{f(x)}{g(x)}$ would oscillate wildly between very large positive and very large negative values, and moreover would be undefined at values of $x$ arbitrarily close to $c$ but not equal to $c$.

So if we are close enough to $c$ but on the right of $c$, then $\frac{f(x)}{g(x)}$ does not change sign. Similar considerations apply to the left of $c$ but close to $c$. However, the sign of $g(x)$ could be different for $x$ close to $c$ and bigger than $c$, and for $x$ close to $c$ and less than $c$,

The author chose to make $g(x)$ positive close enough to $c$ but to the right of $c$. That was an arbitrary choice, negative would have been OK too. Then a minor modification of the argument would show that $\frac{f(x)}{g(x)}\to -\infty$ as $x\to 0^+$.

The reason a $b$ was chosen so that in $(c,b)$ we have $g(x)\gt 0$ is to make sure that we are working in an interval in which the ratio $\frac{f(x)}{g(x)}$ is defined.

Now we look at the first restriction, $\frac{f(c)}{2}\lt f(x)\lt 3\frac{f(c)}{2}$. Only half of that restriction is necessary.

We know that $f(c)\ne 0$. The intuition is that we want to make $\frac{f(x)}{g(x)}$ huge. That will happen since $g(x)$ is approaching $0$, unless $f(x)$ decides to be very close to $0$ also. In that case, we would know nothing about the ratio. So we want to work in an interval such that in that interval $f(x)$ never gets close to $0$. That's why the condition $\frac{f(c)}{2}\lt f(x)$ is needed, to make sure $f(x)$ doesn't get too small.

The inequality $f(x)\lt \frac{3f(c)}{2}$ is not needed (and not used) in the proof. So you are quite right in wondering why it was mentioned. It gets produced automatically when we take care of the $\frac{f(c)}{2}$ part, but still need not have been mentioned.

Answer 2

The idea behind the proof is to bound $f(x)$'s value round a small enough interval contains $c$, such that $f(x)$ deviate at most some amount in this interval. So you can write $M<f(x)<N$ for $x\in I$. Then when we take $f(x)/g(x)$, because $f(x)$ does not change sign at $f(c)$, we have an inequality of type $|f(x)/g(x)|\ge\min(|M/g(x)|,|N/g(x)|)$. Since $M,N$ are bounded, $g(x)$ goes to $0$ when $x$ goes to $c$, the previous inequality implies $f/g$ must be unbounded. We actually know $g(x)$'s behavior around $c$ since $g$ is continuous, so if we approach from either side we would get a vertical asymptote.

For your question (1), imaging we had a discontinuous function like $g(x)=x\sin[1/x]$(with modification $g(0)=0$) such that $g(x)$ changes signs infinitely many times around $x$. Then it is impossible to find the $b$ such that in $c<x<b$ we have $g(x)>0$, and the above proof would collapse. Another exception case is that $g(x)\equiv 0$, but it is also ruled out by the condition in the problem. So the condition given in the problem is strong enough to guarantee $b$'s existence. I believe you can prove this yourself.

Question(2) is hopefully answered above.