Expert: Paul Klarreich - 10/2/2010

Question
Given the graph f(x)=sqr-rootx find a number delta such that

if the absolute value of x-4<d then the absolute value of square root of x-2<.4

I've tried solving for x, but I got the problem wrong, what should I do?

Answer
Questioner:   Rebecca
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Delta, Epsilon
Question:  Given the graph f(x)=sqr-rootx find a number delta such that

if the absolute value of x-4<d then the absolute value of square root of x-2<.4

I've tried solving for x, but I got the problem wrong, what should I do?
.......................
You are trying to prove that

lim[x->4] sqrt(x) = 2.

So you want to show that WHENEVER  | x - 4 | < d, [d is delta] then

| sqrt(x) - 2 | < e.

Of course the 'd' depends on e [e is epsilon].

In this case your  e = 0.4

So you want:

| sqrt(x) - 2 | < 0.4

Do some algebra, such as rationalizing:
| sqrt(x) - 2 || sqrt(x) + 2 |
------------------------------ < 0.4
       | sqrt(x) + 2 |


     | x - 4 |
------------------------------ < 0.4
 | sqrt(x) + 2 |

Now when x is near 4,  | sqrt(x) + 2 | is near | sqrt(4) + 2 |,
which is near   | 2 + 2 | = 4.  The worst would be x = 0, so we will use that:

| x - 4 |
-------------- < 0.4
 | 0 + 2 |


| x - 4 |
-------------- < 0.4
   2

| x - 4 | < 0.8

That  0.8 is your delta.