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. $$\lim_{x->5}, 1/x= 1/5, \epsilon=.05$$. That implies the following $$|\frac{1}{x}-\frac{1}{5}|< \epsilon \mbox{ if }|x-5|<\delta$$. Which implies $$|\frac{1}{x}-\frac{1}{5}|< .05 \\$$. Which gives $$.15< \frac{1}{x} < .25$$. Which gives $$6\frac{2}{6}> x > 4 \mbox{ therefore } 1\frac{2}{3} < x-5< -1$$. 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]

$$6\frac{2}{6}> x > 4 \mbox{ therefore } 1\frac{2}{3} < x-5< -1$$.

Here is your mistake. The condition is supposed to be $$|x-5|<\delta$$.

 

[John O' Meara]

Do you mean that the arrows are the wrong way around $$6\frac{2}{3}>x>4 \mbox{ therefore } 1\frac{2}{3}>x-5>-1$$.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 $$\lim_{\frac{1}{x}->5}$$. Thanks.