# Finding the largest value of delta greater than 0

1. Sep 10, 2012

### zurburgk

1. The problem statement, all variables and given/known data
For f(x) = sqrt(27-x), L=4, x_0 = 11 epsilon = 1, find the largest value of delta > 0 in the formal definition of a limit which ensures that |f(x) - L| < epsilon

2. Relevant equations

the formal def. of a limit:
lim x->x_0 F(x) = L if, for every number epsilon > 0, there exists a corresponding number delta>0 such that
0<|x-x_0| < delta implies |f(x) - L| < epsilon

3. The attempt at a solution

I've tried plugging multiple different things in to the equations, but i, sadly, have no idea what i'm doing.

0 < |x-11| < epsilon and |sqrt(27-x) - 4| < 1.

Attempting to solve the second equation i get x>2. If i plug that into the first equation i got -9. But that is definitely not the answer. I just don't know what all these numbers mean.

2. Sep 10, 2012

### SammyS

Staff Emeritus
How about solving $\displaystyle |\sqrt{27-x\ } - 4| = 1$ for x, to start things off ?

3. Sep 10, 2012

### zurburgk

Do you mind if i walk you through my steps, if i make a mistake it should be easier to pinpoint the area :)

So |sqrt(27-x)-4| < 1
- add 4 to both sides

|sqrt(27-x)| < 5
-get rid of the square root

|27-x|< 25
-subtract 27 from each side

|-x| < -2

- and here is where i'm having my first problem. Do i multiply both sides by a negative allowing x to be a positive variable? Or do the absolute value signs take care of the negative variable?

4. Sep 10, 2012

### SammyS

Staff Emeritus
Let me repeat,
Solve $\displaystyle |\sqrt{27-x\ } - 4| = 1$ for x.​

But if you insist, you can solve the inequality. Just be careful. That is trickier.

At any rate:
$\displaystyle |\sqrt{27-x\ } - 4|+4\ne |\sqrt{27-x\ }|$​

If $\displaystyle |\sqrt{27-x\ } - 4| < 1$

then $\displaystyle -1<\sqrt{27-x\ } - 4 < 1\ .$

Can you continue from there?