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?
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$\begingroup$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):
- Assume the limit exists, and is some real number $L\in \Bbb R$
- 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.
$\endgroup$ 7 $\begingroup$Because then the value $|f(x)|$ becomes bigger than any $M\in\mathbb R$, therefore no value can be the limit.
$\endgroup$ $\begingroup$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 $\lim_{x\to a} p(x)\neq 0$.
$\endgroup$ 2 $\begingroup$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.
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