# Homework Help: Limit prooving question

1. Jan 6, 2009

### transgalactic

the question and where i got stuck in this link:

http://img187.imageshack.us/img187/8868/44429492ix7.gif [Broken]

Last edited by a moderator: May 3, 2017
2. Jan 6, 2009

### HallsofIvy

You have a proof that limit, as x goes to 5 of x2= 25. For that you must have $|x^2- 25|< \epsilon$ which leads to $|(x- 5)(x+ 5)|< \epsilon$
Since you want something of the form $|x- 5|< \delta$, is is the "x+ 5" term you want to deal with. You cannot write $|x- 5|< \epsilon/(x+5)$ because the right side cannot depend on x. What you can do is find an upper bound: if |x+5|< M, then 1/|x+5|> 1/M so $\epsilon/|x+5|> \epsilon/M$. If, then $|x-5|< \epsilon/M$ which is itself less than $\epsilon/|x+5|$ it is certainly less than $\epsilon/(x+5)$

What is an upper bound on |x+5|? First you must understand that there are an infinite number of such "upper bounds". If a is an upper bound then any number larger than a is still an upperbound. We want x close to 5. Just because "1" is an easy number, lets try |x-5|< 1. That says that -1< x- 5< 1 and, adding 10 to each part 9< x+5< 11. Since 9< x+ 5, it is obviously positive so |x+5|= x+ 5< 11. I chose "10" to add, of course, because x-5+ 10= x+ 5.

That's where those numbers come from. We could as easily have decided that |x-5|< 2 is "close enough". Then we would have -2< x- 5< 2 and so, adding 10 again, 8< x+ 5< 12 and now we "12" as an upper bound on |x+5| rather than 11.

Or we could decide we wanted to be |x-5|< 1/2. -1/2< x- 5< 1/2. Again adding 10, 9.5< x+ 5< 10.5 so that 10.5 is an upper bound. In fact, any number greater than 10 could be used as an upper bound.

3. Jan 6, 2009

### transgalactic

Thanks i understand that

4. Jan 6, 2009

### transgalactic

i understand that our upper bound is 11

so inorder for this to work we need only

$\delta = \frac{\epsilon }{11}$

why to use minimum??

why they are taking the solution as the minimal between 1 and the other value??

$\delta =\min \left\{ 1,\frac{\epsilon }{11} \right\}.\quad\blacksquare$

5. Jan 7, 2009

### HallsofIvy

We have shown that in order that |x- L|< $\epsilon$, we must have |x- a|, $\epsilon/11$. But in order to have that we must have |x-a|<1. That is, they must both be true.

If |x-a| is less than the smaller of the two, it is less than both.