1. ## Epsilon Delta Problem

How to I find the delta that works for epsilon < 0.006?

i know that abs(f(x)-L) = abs(1/x-1/6) = abs(6-x / 6x) = abs(6-x) abs(1/6x)

But how can i simply it further to match abs(x-6) and what i should bound abs(1/6x) to?

2. ## Re: Epsilon Delta Problem

Originally Posted by lc99

How to I find the delta that works for epsilon < 0.006?
i know that abs(f(x)-L) = abs(1/x-1/6) = abs(6-x / 6x) = abs(6-x) abs(1/6x)
Note that $\left|\dfrac{6-x}{6x}\right|=\dfrac{|x-6|}{6|x|}$ If $|x-6|<1 \Rightarrow~5<x<7$ so that $\dfrac{1}{|x|}<\dfrac{1}{5}$.
What if $\delta=0.003~??$

3. ## Re: Epsilon Delta Problem

You are over-complicating it:
$$\begin{array}{r r c c c l l} \left| \frac1x - \frac16 \right| < \frac6{1000} \implies & -\frac{6}{1000} & < & \frac1x - \frac16 & < & \frac6{1000} \\ & \frac16 - \frac6{1000} & < & \frac1x & < & \frac16 + \frac6{1000} \\ & \frac{964}{6000} & < & \frac1x & < & \frac{1036}{6000} \\ & \frac{6000}{1036} & < & x & < & \frac{6000}{964} \\ & \frac{6216 - 216}{1036} & < & x & < & \frac{5784 + 216}{964} \\ & 6 - \frac{216}{1036} & < & x & < & 6 + \frac{216}{964} \\ & -\frac{216}{1036} & < & x - 6 & < & \frac{216}{964} & \implies \left|x -6\right| < \frac{216}{1036} \end{array}$$

4. ## Re: Epsilon Delta Problem

Originally Posted by lc99

How to I find the delta that works for epsilon < 0.006?

i know that abs(f(x)-L) = abs(1/x-1/6) = abs(6-x / 6x) = abs(6-x) abs(1/6x)

But how can i simply it further to match abs(x-6) and what i should bound abs(1/6x) to?
\displaystyle \begin{align*} \left| \frac{1}{x} - \frac{1}{6} \right| &< \frac{3}{500} \\ \left| \frac{6 - x}{6\,x} \right| &< \frac{3}{500} \\ \frac{\left| 6 - x \right| }{\left| 6\,x \right| } &< \frac{3}{500} \\ \frac{1}{6} \cdot \frac{1}{\left| x \right| } \cdot \left| x - 6 \right| &< \frac{3}{500} \\ \frac{1}{ \left| x \right| } \cdot \left| x - 6 \right| &< \frac{9}{250} \\ \left| x - 6 \right| &< \frac{9}{250} \left| x \right| \end{align*}

Now if we restrict \displaystyle \begin{align*} \left| x - 6 \right| < 1 \end{align*} say, then that means

\displaystyle \begin{align*} \left| x \right| - \left| 6 \right| \leq \left| x - 6 \right| &< 1 \textrm{ by the Reverse Triangle Inequality } \\ \left| x \right| - 6 &< 1 \\ \left| x \right| &< 7 \\ \frac{9}{250} \left| x \right| &< \frac{9}{250} \cdot 7 \\ \frac{9}{250} \left| x \right| &< \frac{63}{250} \end{align*}

So since \displaystyle \begin{align*} \left| x - 6 \right| < \frac{9}{250} \left| x \right| < \frac{63}{250} \end{align*}, that means we can set \displaystyle \begin{align*} \delta = \textrm{min}\,\left\{ 1 , \frac{63}{250} \right\} \end{align*}.

5. ## Re: Epsilon Delta Problem

That isn't what the question asks. Moreover
Originally Posted by Prove It
$\delta = \textrm{min}\,\left\{ 1 , \frac{63}{250} \right\}$
Can be written more easily as $\delta = \frac{63}{250}$

6. ## Re: Epsilon Delta Problem

Originally Posted by Archie
That isn't what the question asks. Moreover

Can be written more easily as $\delta = \frac{63}{250}$
It's not?

i know that abs(f(x)-L) = abs(1/x-1/6) = abs(6-x / 6x) = abs(6-x) abs(1/6x)

But how can i simply it further to match abs(x-6) and what i should bound abs(1/6x) to?
Seems to be EXACTLY what was being asked.

As for how the delta can be written more easily, that is true, but I wanted to leave it in the form that one would have to leave it in more general epsilon-delta proofs.