# Why doesn't a limit exist if you have 0 in the denominator?

Say I want to find the limit of $f(x) = \frac{p(x)}{q(x)}$ at some 'a'. I've been told the limit doesn't exist if ever the denominator q(x) went to zero but not the numerator. Why is this?

• This is because your fraction doesn't approach any number at all; it just increases without bound – Zachary Nov 13 '17 at 12:01
• Because $0/c$ is just $0$, but $c/0$ is not defined in most contexts. Does not have much to do with limits. Or do you ask why $c/0$ is bad? – M. Winter Nov 13 '17 at 13:21
• @M.Winter It has everything to do with limits. Taking limits is not the same as inserting $x = a$ and evaluating. Even more so if you allow discontinuous functions. – Arthur Nov 13 '17 at 13:24
• @Arthur Well, ok. It has something to so with limits. But the problem arises before that when one asks why it is bad to divide by zero. – M. Winter Nov 13 '17 at 13:31
• @M.Winter No, that's not where the problem arises. The definition of limits specifically avoids the exact point of interest $a$, and asks what happens on every other point in the immediate vicinity of $a$t. And at these other points in the vicinity, $\frac{p(x)}{q(x)}$ is (at least in most problems given) an unproblematic fraction. – Arthur Nov 13 '17 at 13:35

Here are the steps that I would take to prove it, under the assumption that $p$ and $q$ are continuous (without that assumption, or something very like it, there really isn't much you can do in general):

1. Assume the limit exists, and is some real number $L\in \Bbb R$
2. Use the following known facts in concert to derive a contradiction:
• The definition of $\lim_{x\to a}\frac{p(x)}{q(x)} = L$
• $q(a) = 0 \neq p(a)$
• $p$ and $q$ are continuous

Edit: Thorough working out:

We ultimately want to disprove that $\lim_{x\to a}\frac{p(x)}{q(x)} = L$, so we just need to find a single $\epsilon>0$ that makes a contradiction. I pick $1$, because I like it (and because I actually know that they will all fail, so it doesn't matter which one I pick, so I go for one that is easy to work with). Since we assumed that the limit existed, that must mean that there is a $\delta>0$ that fulfills the definition $\lim_{x\to a}\frac{p(x)}{q(x)} = L$ for this specific value of $\epsilon$. In other words, for any $x\in (a-\delta, a+\delta)\setminus \{a\}$, we have $$\left|\frac{p(x)}{q(x)} - L\right|<1\\ \left|\frac{p(x) - Lq(x)}{q(x)}\right|<1\\ \frac{|p(x)-Lq(x)|}{|q(x)|}<1\\ |p(x) - Lq(x)| < |q(x)|$$ Now let's use that $p$ and $q$ are continuous. $p$ being continuous means that there is a $\delta_p>0$ such that for any $x\in (a-\delta_p, a+\delta_p)$, we have $|p(x) - p(a)|<\frac{|p(a)|}2$ (here we use that $p(a)\neq 0$). This, in turn, means that $|p(x)|>\frac{|p(a)|}2$.

Similarily, we know that there is some $\delta_q>0$, such that for any $x\in (a-\delta_q, a+\delta_q)$ we have $|q(x)-q(a)| = |q(x)| <\frac{|p(a)|}{2(|L| + 1)}$ (here we also use that $p(a) \neq 0$, along with $q(a) = 0$). This, in turn, means that $(|L| + 1)|q(x)|<\frac{|p(a)|}{2}$.

If we now assume that $x\in (a-\delta_p, a+\delta_p)\cap (a-\delta_q, a+\delta_q)$, we can chain these two implications together to get $$|Lq(x)| + |q(x)| < \frac{|p(a)|}{2} < p(x)\\ |q(x)|<|p(x)| - |Lq(x)|\\ |q(x)|<|p(x) - Lq(x)|$$ Now, let $\delta_f = \min(\delta, \delta_p, \delta_q)$, and pick an $x\in (a-\delta_f, a+\delta_f)\setminus\{a\}$, which is possible since $\delta_f>0$. For such an $x$, we have both $|q(x)|<|p(x) - Lq(x)|$ and $|p(x) - Lq(x)| < |q(x)|$, which is a contradiction.

• how to deduce contradiction? my attempt:$|\frac{p(x)}{q(x)} - L| < \epsilon \implies q(x)(L-\epsilon ) < p(x) <q(x)(L+\epsilon)$ when ever $x \in (a-\delta , a+\delta)$ as $x \to a$ we get in above $p(a) \to 0$ (by squeeze theorem) – Magneto Nov 13 '17 at 12:06
• okay, let the limit be L. you have limit p(x) *limit 1/q(x) = L. This is because they're continous. limit 1/q(x) = L/limit(p(x)). But first limit doesn't exist therefore limit doesn't exist. Does this look alright? – Vrisk Nov 13 '17 at 12:11
• @Vrisk I added a (hopefully rather exhaustive) proof using the definitions directly. – Arthur Nov 13 '17 at 13:05
• You assumed p and q are polynomials. -1 – windircurse Nov 13 '17 at 13:12
• @windircurse There. I added a small clause, and changed three words in the entire answer. Don't know how such an insignificant mistake earned a -1, but at least you told me (which is better than many downvotes I get). – Arthur Nov 13 '17 at 13:15

Because then the value $|f(x)|$ becomes bigger than any $M\in\mathbb R$, therefore no value can be the limit.

You can appeal to the definition of limit, or you can use the product law and prove this by contradiction:

Suppose that $\lim_{x\to a} p(x) = b \neq 0$, and $\lim_{x\to a} q(x) = 0$. Then $\lim_{x\to a} \frac{p(x)}{q(x)}$ does not exist.

Proof Suppose for the sake of contradiction that $\lim_{x\to a} \frac{p(x)}{q(x)} = L$. Then $$\lim_{x\to a} p(x) = \lim_{x\to a} \frac{p(x)}{q(x)}\cdot q(x) = \lim_{x\to a} \frac{p(x)}{q(x)} \cdot \lim_{x\to a} q(x) = L \cdot 0 = 0$$ But this contradicts the assumption that $\lim_{x\to a} p(x)\neq 0$.

• isn't there division by 0 in your first step - more precisely 0/0 != 1 – Vrisk Nov 13 '17 at 14:27
• nvm, i got it. thanks, really neat and simple +1 – Vrisk Nov 13 '17 at 14:44

Try it yourself (and use a calculator if you want). Choose a small number $x$, e.g. $x=0.001$, and compute both

$$\frac x1,\qquad\qquad \frac 1x.$$

Now choose a smaller number for $x$, e.g. $x=0.00001$, and compute again. Do this again and again with smaller numbers. Can you see how $x/1$ becomes smaller and smaller, almost zero? Can you see how this is not the case for $1/x$? It gets bigger and bigger instead and there is no number for which we can say that the sequence approches it (and no, $\infty$ is no number).

It does not really matter what is on the other side of the quotient. When only the numerator vanishes, everything tends to zero. When only the denominator vanishes, everything tends to explode.