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

[tex]6\frac{2}{6}> x > 4 \mbox{ therefore } 1\frac{2}{3} < x-5< -1[/tex].

Here is your mistake. The condition is supposed to be [tex]|x-5|<\delta[/tex].

 

[John O' Meara]

Do you mean that the arrows are the wrong way around [tex] 6\frac{2}{3}>x>4 \mbox{ therefore } 1\frac{2}{3}>x-5>-1[/tex].The actual answer to [tex]\delta = \frac{1}{505} [/tex]. 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 [tex] \lim_{\frac{1}{x}->5} [/tex]. Thanks.