Algebraic Properties of Limits: Quotient and Reciprocals

As in our proof of the product rule we shall assume that $\lim_{x \to x_{0}}f(x) = A$ and $\lim_{x \to x_{0}}g(x) = B$ .

Initially we want to prove the reciprocal rule, which states:

$\lim_{x \to x_{0}}\dfrac{1}{f(x)} = \dfrac{1}{A}$

Proof of Reciprocal Law:

To prove this we shall have to show that for some arbitrary $\epsilon$ we have:

$|\dfrac{1}{f(x)} - \dfrac{1}{A}| < \epsilon$ whenever $0< |x - x_{0}| < \delta$

Hence we take

$|\dfrac{1}{f(x)} - \dfrac{1}{A}| \\ = |\dfrac{1A - 1f(x)}{f(x)A}| = |\dfrac{A - f(x)}{f(x)A}| = |\dfrac{1}{f(x)A}||(A-f(x))|$

If we do a little rewriting we come to see that:

$|\dfrac{1}{f(x)A}||(A-f(x))| = |\dfrac{1}{f(x)} \dfrac{1}{A}| |(f(x) - A)|$ Call this (*P-step*)

So now what? Well we need to use the facts we have, specifically the fact that the limit A exists for the function f(x). Since the function $f(x)$ occurs twice in (*P-step*) we will have to find two restrictions. Note that while $\lim_{x \to x_{0}}f(x) = A$ implies the existence of a $\delta_{1}$ , it also ensures the existence of a $\delta_{2}$ , since whenever we have:

$|f(x) - A| < \epsilon_{2}$ if $0 < |x - x_{0}| < \delta_{1}$

we have

$|A - f(x)| < \epsilon_{1}$ if $0 < |x -x_{0}| < \delta_{2}$

since the absolute values of the distances are equal. Let $\delta = min( \delta_{1}, \delta_{2})$

We define $\epsilon_{1} = \dfrac{|A|}{2}$ and $\epsilon_{2} = \dfrac{|A|^{2}}{2}*\dfrac{\epsilon}{1}$ . The reason for this specification will become apparent as we see that (a) these $\epsilon - neighborhoods$ allow us to place upper bounds on the factors in (*P-step*) and (b) cancel in such a way that allows us to prove the reciprocal rule.

First we show (a): Note that

$|A| = |A - f(x) +f(x)| \leq |A - f(x)|+|f(x)|$

by the triangle inequality, so by our definition of $\epsilon_{1}$ and we have:

$|A| \leq |A - f(x)|+|f(x)|< \epsilon_{1}+|f(x)| = \dfrac{|A|}{2} + |f(x)|$

From which it follows that:

$\dfrac{|A|}{2} < f(x) = \dfrac{1}{f(x)} < \dfrac{2}{|A|}$

since $|A| < \dfrac{|A|}{2} +f(x)$ and we can subtract $\dfrac{A}{2}$ from both sides. Finally we substitute this discovered restriction into (*P-step*), we get:

$|\dfrac{1}{f(x)}\dfrac{1}{A}||(A-f(x))| < \dfrac{2}{|A|} \dfrac{1}{A} |(f(x) - A)| = \dfrac{2}{A^{2}}|(f(x) -A)|$

We need now to specify a restriction on |f(x) – A| such that our restriction will cancel with $\dfrac{2}{A^{2}}$ to create $\epsilon$ . But we have already defined a second such restriction in $\epsilon_{2}$ ! From our initial calculations and our observed restrictions, we see (b) that:

$|\dfrac{1}{f(x)} - \dfrac{1}{A}| = |\dfrac{1}{f(x)}\dfrac{1}{A}||(A-f(x))|$ < $\dfrac{2}{A^{2}} \dfrac{A^{2}}{2}*\dfrac{\epsilon}{1}$

and of course:

$\dfrac{2}{A^{2}} \dfrac{A^{2}}{2}*\dfrac{\epsilon}{1} = \dfrac{\epsilon}{1} = \epsilon$

Hence $|\dfrac{1}{f(x)} - \dfrac{1}{A}| < \epsilon$ as desired. This concludes the proof of the reciprocal law.

Proof of Quotient Rule:

Now the proof of the quotient rule is easy. We need to show that $\lim_{x \to x_{0}}[ \dfrac{f(x)}{g(x)}]$ = $\dfrac{\lim_{x\to x_0}f(x)}{\lim_{x \to x_0}g(x)}$ = $\dfrac{A}{B}$ provided $\lim{ x \to x_{0}}[g(x)] \neq 0.$ To show this we have:

$\lim_{x \to x_{0}}[ \dfrac{f(x)}{g(x)}]$ = $\lim_{x \to x_{0}}[ (f(x)\dfrac{1}{g(x)}]$

But then by the product rule we have:

$\lim_{x \to x_{0}}[(f(x)]$ $\lim_{x \to x_{0}}[\dfrac{1}{g(x)}]$

from which we know, by an application of the reciprocal rule

$\lim_{x \to x_{0}}[ \dfrac{f(x)}{g(x)}]$ = $A(\dfrac{1}{B}) = \dfrac{A}{B}.$

…and we are done, as this completes this proof of the quotient rule.

19Jan2014

Algebraic Properties of Limits: Product Rule

Posted in Mathematics, Things that are true by nathanielforde

Assume that $\lim_{x \to x_{0}}[f(x)] = A$ and $\lim_{x \to x_{0}}[g(x)] = B$ . We wish to show that the $\lim$ distributes over the multiplication operation. Or more precisely we will prove that

$\lim_{x \to x_{0}}[f(x)g(x)] = \lim_{x \to x_{0}}[f(x)] \lim_{x \to x_{0}}[g(x)] = AB$

This proof is difficult but it effectively demonstrate the kinds of thinking often required by $\epsilon \delta$ proofs.

Proof #1: For this proof we will use the triangle inequality like we did for the Sum and Difference rules. See here.

Suppose $\epsilon > 0$ . Observe that

$|f(x)g(x) - AB| \\ = |f(x)g(x) - A(g(x)) + A(g(x)) - AB| \\ = | g(x) (f(x) - A) + A(g(x) - B)| \\ \leq |g(x)| |(f(x) - A)| + |A||(g(x) - B)|$

In this last clause we have used the triangle inequality, but more than that we have made our task easier for we now have a lower bound for the value of $\epsilon$ in the sum of two terms. By assumption we know there are two values $\delta_{1} \delta_{2}$ . We want to have each such value be lower than $\epsilon/2$ .

We will focus on the righthand term first in the above equation. So because we know that $\lim_{x \to x_{0}}[g(x)] = B$ , we know there is a $\delta_{2}$ to be found – the smaller the better – which corresponds to a specific $\epsilon_{2}$ such that:

$|g(x) - B| < \epsilon_{2}$ whenever $0 < |x - x_0| < \delta_{2}$

Given our intention we let $\epsilon_{2} = \frac{\epsilon}{2} * \frac{1}{(1+|A|)} < \frac{\epsilon}{2}.$

Again since $\lim_{x \to x_{0}}[f(x)] = A$ we know there is a $\delta_{1}$ to be found such that:

$|f(x) - A| < \epsilon_{2}$ whenever $0 < |x - x_0| < \delta_{1}$

with $\epsilon_{1} = \frac{\epsilon}{1} * \frac{1}{(1+|B|)} < \frac{\epsilon}{2}.$

In addition we shall specify that there is at most an “ $\epsilon-neighborhood$ of depth $\pm1$ such that:

$|g(x) - B| < 1$ whenever $0 < |x - x_{0}| < \delta_{3}$

then adding |B| to both sides of the left inequality we get:

$|g(x) - B + B| < 1 + |B| = |g(x)| < 1+ |B|$

So let our $\delta$ = min( $\delta_{1}, \delta_{2}, \delta_{3}$ ). This covers all cases.

Now recall our triangular inequality, from above:

$|f(x)g(x) - AB| \leq$ $|g(x)| |(f(x) - A)| + |A||(g(x) - B)|$

from this we know that

$|f(x)g(x) - AB| <$ $|g(x)| |(f(x) - A)| + |A+1||(g(x) - B)|$

Now we substitute what we know about $\epsilon_{1}, \epsilon_{2}$ to get:

$|f(x)g(x) - AB| <$ $|g(x)| \frac{\epsilon}{2(B+1)} + |A+1|\frac{\epsilon}{2(A+1)}$

Finally, we substitute our discovery about the upper bound for |g(x)| to see:

$|f(x)g(x) - AB| <$ $|B+1| \frac{\epsilon}{2(B+1)} + |A+1|\frac{\epsilon}{2(A+1)}$

Cancelling, it becomes clear:

$|f(x)g(x) - AB| <$ $\frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

as desired. This completes the proof since we have shown that $|f(x)g(x) - AB| < \epsilon$ when $0 < |x - x_{0}| < \delta$ .

18Jan2014

Algebraic Properties of Limits: Easy Pieces

Posted in Mathematics, Things that are true by nathanielforde

In the last post here, we proved that $\lim$ distributes over addition and subtraction. We shall now prove a few easy pieces that will ultimately help us in our attempt to show how $\lim$ behaves when interacting with multiplication and division. Assume $\lim_{x \to x_0}[f(x)] = A$ , we will prove…

$\lim_{x \to x_0}[c] = c$ , for some constant c.
$\lim_{x \to x_0}[x] = x_{0}$
$\lim_{x \to x_0}[cf(x)]$ = $c \lim_{x \to x_0}[f(x)] = cA.$
$\lim_{x \to x_0}[f(x) - A] = 0$

We prove (1) first.

Suppose $\epsilon > 0$ . We want to show that where a function $f(x) = c$ for all x, then we can find a $\delta$ such that $|f(x) - c| < \epsilon$ while $0< |x - x_{0}| < \delta.$ For this, pick any $\delta$ whatsoever! Since we already know that $|c - c | = 0 < \epsilon$ we are done.

Now we prove (2):

Again we have to specify a function $f(x) = x$ for all x, now suppose $\epsilon > 0$ and let $\delta = \epsilon$ , so that once we assume $0 < |x - x_{0}| < \delta$ , then it’s obvious that $|f(x) - x_{0}| = |x - x_{0}| < \delta = \epsilon.$ As desired.

…and finally (3):

There are two cases here (i) where $c = 0$ and (ii) where $c \neq 0.$ The first case easy since if c = 0, then $\lim_{x \to x_{0}}[cf(x)]$ = $\lim_{x \to x_{0}}[0f(x)]$ = $\lim_{x \to x_{0}}[0]$ , but then by (1) we know that $\lim_{x \to x_{0}}[0] = 0 = 0\lim_{x \to x_0}[f(x)].$ So we’re done with this case.

Now consider the second case (ii). Let $\epsilon > 0$ . By assumption we know that $\lim_{x \to x_0}[f(x)] = A$ , so there is a number $\delta_{1}$ , such that whenever $0 < |x - x_{0}| < \delta_{1}$ we have that $|f(x) - A| < \epsilon/|c|$ . Let $\delta = \delta_{1}$ , remember we need to prove that