Expert: Paul Klarreich - 2/23/2008

Question
lim (x+6)/x = 4  as x->2

I find delta=epsilon/3

There is an online applet at
www.scottsara.org/applets/calculus/epsilondelta.html

But my answer doesn't look correct.
Whats WRONG ?

Any light you can shed would be appreciated
Thanks

Answer
Questioner:   Mark
Category:  Calculus
Private:  No
 
Subject:  Delta epsilon Method Precise Limits
Question:  lim (x+6)/x = 4  as x->2

I find delta=epsilon/3

There is an online applet at
www.scottsara.org/applets/calculus/epsilondelta.html

>> THIS LINK DID NOT WORK.

But my answer doesn't look correct.
Whats WRONG ?

Any light you can shed would be appreciated
Thanks
..................................
Hi, Mark,

I have to complain to you about something.

In my INSTRUCTIONS to questioners, I ask that you send me all the work you did.  Then I can answer your question of 'Whats WRONG?'  I cannot tell what you did wrong if I can't see it.

In proving that:

lim   (x+6)/x = 4
x->2

You would want (I don't make epsilons and deltas, just e's and d's):

| (x+6)/x - 4 | < e, whenever  | x - 2 | < d

| (x+6 - 4x)/x | < e

| (6 - 3x)/x | < e

| (2 - x)/x | < e/3

Now you must make an assumption about x.  
Suppose you assume that x >= 1, since x is supposed to be near 2

If  x >= 1, then 1/x <= 1, and

| (2 - x)/x | <= |2 - x|

So your inequality becomes:

| 2 - x | < e/3, which is the same as:

| x - 2 | < e/3

So you can take  d = e/3, AS YOU DID.

But the assumption of  x >= 1 is not the only one, so there can be other correct answers.