These four problems are 'bonus' problems for extra credit, because of how hard they are.


Well, hard in the sense that you don't see the way to solve them right away. And our professor didn't do any examples of them


If anyone can help me with these, even just getting started, I would be grateful.


Here they are:


Prove each of the following limits using the Epsilon - Delta definition of a limit. Lim (as x-->a) of f(x) = L if and only if for every number Epsilon > 0 there exists a number Delta such that 0 < |x-a| < Delta then |f(x) - L| < Epsilon



1) Lim (as x 3) of 4 / (x-1) = 2



2) Lim (as x 4) of Sqrt(x+5) = 3



3) Lim (as x 2) of (x+3) / (x-1) = 5



4) Lim (as x 1) of 1 / Sqrt(5-x) = 1/2

This post was edited by Eep on Sep 26 2011 10:36pm

too hard for jsp :[

Just because your professor didn't do examples of them doesn't mean they're not a standard type of problem - epsilon-delta arguments crop up all the time in multiple areas of mathematics. Try Google or Wikipedia for some examples.

I'd have a crack at them myself but I'm a bit rusty...

too boring for jsp...

This is true as well. It's a really tedious thing to do unless you've really been practising them. The basic format is always the same though, so go get some examples and follow them through as carefully and logically as you can. It's a very important part of calculus, conceptually.

I am not sure if you guys are understanding my dilemma


I know how to prove a limit using the definition


but these 4 examples are a little more unorthodox than other ones I've seen


if I could get a hint as to how to solve them, like just to get started, that would be great

as in, the |f(x)-L| < Epsilon part




This post was edited by Eep on Sep 29 2011 10:21pm

You could probably divide domain into parts and take out absolute value. Work from there and you should get answers.

Usually offering to pay people helps.

|f(x) - L| < Epsilon

1. |(4/(x-1))- 2 | < e

|(4/(x-1)|<e+2
|(x-1)/4|<1/(e+2)
|(x-1)|<4/(e+2)
|x-3|<(4/(e+2))-2


2.Lim (as x 4) of Sqrt(x+5) = 3

|sqrt(x+5)-3|<e
|sqrt(x+5)|<(e+3)
|x+5|<(e+3)^2
|x-4|<(e+3)^2 -9

should be able to go on from there

This post was edited by saber_x3 on Oct 1 2011 09:35am

|(4/(x-1))- 2 | < e

|(4/(x-1)|<e+2

Can't do that., This is only true for x > 3.