How do you know what delta to choose

[blackblanx]

Find the delta for the given epsilon. lim (1/x) =1 epsilon=.07
x->1

Homework Equations

The Attempt at a Solution

I got to here .07526 >x-1> -.06542 so what one is me delta??

 

[aPhilosopher]

I got to run, but it will help whoever helps you if you say how you got there.

 

[Mark44]

Find the delta for the given epsilon. lim (1/x) =1 epsilon=.07
x->1

Homework Equations

The Attempt at a Solution

I got to here .07526 >x-1> -.06542 so what one is me delta??

You want a positive number for delta.

 

[blackblanx]

You want a positive number for delta.

I got the problem from http://archives.math.utk.edu/visual.calculus/1/definition.8/index.html and they say that delta is .06542 but they are unclear of how they decided that. How did they come to that conclusion?

 

[snipez90]

Let \varepsilon > 0. Assume the existence of a \delta > 0 such that 0 < |x-1| < \delta. Then

\left|\frac{1}{x} - 1\right| = |x-1|\cdot\frac{1}{|x|} .

The |x-1| term already has a bound, so we need some bound on 1/|x|. Suppose that \delta \leq 1 so that
|1| - |x| \leq |1-x| < \delta \Rightarrow |x| > 1 - \delta \Rightarrow \frac{1}{|x|} < \frac{1}{1-\delta}.

This implies that

\left|\frac{1}{x} - 1\right| = |x-1|\cdot\frac{1}{|x|} < \frac{\delta}{1 - \delta}.

Requiring this last expression on the right to be \leq \epsilon, we find that setting

\delta = \min(1, \frac{\varepsilon}{\varepsilon + 1})\mbox{ ensures that } \left|\frac{1}{x} - 1\right| < \varepsilon.

Note: We chose delta to be less than or equal to 1 to make it easier to find a bound for 1/|x|, but we could have picked a number smaller than 1. Any number a > 1 would not work, since this would imply that |x-1| < a which would mean that x is in (1-a, 1+a) and since 1-a < 0 < 1 + a, division by zero could occur.

Note: Plugging in epsilon = .07 gives the delta you were looking for.

 

[blackblanx]

Oh, I see thank you snipez.

 

[blackblanx]

Let \varepsilon &gt; 0. Assume the existence of a \delta &gt; 0 such that 0 &lt; |x-1| &lt; \delta. Then

\left|\frac{1}{x} - 1\right| = |x-1|\cdot\frac{1}{|x|} .

The |x-1| term already has a bound, so we need some bound on 1/|x|. Suppose that \delta \leq 1 so that
|1| - |x| \leq |1-x| &lt; \delta \Rightarrow |x| &gt; 1 - \delta \Rightarrow \frac{1}{|x|} &lt; \frac{1}{1-\delta}.

This implies that

\left|\frac{1}{x} - 1\right| = |x-1|\cdot\frac{1}{|x|} &lt; \frac{\delta}{1 - \delta}.

Requiring this last expression on the right to be \leq \epsilon, we find that setting

\delta = \min(1, \frac{\varepsilon}{\varepsilon + 1})\mbox{ ensures that } \left|\frac{1}{x} - 1\right| &lt; \varepsilon.

Note: We chose delta to be less than or equal to 1 to make it easier to find a bound for 1/|x|, but we could have picked a number smaller than 1. Any number a > 1 would not work, since this would imply that |x-1| < a which would mean that x is in (1-a, 1+a) and since 1-a < 0 < 1 + a, division by zero could occur.

Note: Plugging in epsilon = .07 gives the delta you were looking for.

but can you explain your steps a little more please, I've never seen it done like that before. up at the top how did X-1<delta turn into l1l-lxl<=l1-xl<=delta