Limits, find delta given epsilon

[John O' Meara]

A positive number epsilon (e) and a limit L of a function f at a are given. Find delta such that |f(x)-L|< epsilon if 0 < |x-a| < delta. \lim_{x-&gt;5}, 1/x= 1/5, \epsilon=.05. That implies the following |\frac{1}{x}-\frac{1}{5}|&lt; \epsilon \mbox{ if }|x-5|&lt;\delta. Which implies |\frac{1}{x}-\frac{1}{5}|&lt; .05 \\. Which gives .15&lt; \frac{1}{x} &lt; .25. Which gives 6\frac{2}{6}&gt; x &gt; 4 \mbox{ therefore } 1\frac{2}{3} &lt; x-5&lt; -1. Which does not give the correct delta. I maybe rusty on algebra, as I am studying on my own. Could someone show me how to do it correctly. Thanks for the help.

 

[LeonhardEuler]

6\frac{2}{6}&gt; x &gt; 4 \mbox{ therefore } 1\frac{2}{3} &lt; x-5&lt; -1.

Here is your mistake. The condition is supposed to be |x-5|&lt;\delta.

 

[John O' Meara]

Do you mean that the arrows are the wrong way around 6\frac{2}{3}&gt;x&gt;4 \mbox{ therefore } 1\frac{2}{3}&gt;x-5&gt;-1.The actual answer to \delta = \frac{1}{505}. But I cannot get it.

 

[LeonhardEuler]

Are you sure you copied the question correctly? If you use 1/x ->5 as x->1/5 instead of the other way around you get that answer.

 

[John O' Meara]

I checked out the question and I have copied it correctly, they must have meant \lim_{\frac{1}{x}-&gt;5}. Thanks.