# I will use a 'g' for delta and a small 'e' for episilon to find delta

• Sep 29th 2009, 04:56 AM
bosmith
I will use a 'g' for delta and a small 'e' for episilon to find delta
I must find the value of 'g'>0 such that |f(x) - L| < 'e' when 0 < |x-a| < 'g' lim x-> 9 (square root 3) = 3, 'e'= 0.01. I think the function of |x^(1/2 -3|< 0.01 is the proper way to set up this problem. I am having trouble at this point Generally for all of the problems I have done so far included |x-9| as x -> 9. Then I find -1 < |x-9| < 1 -> which results in 8< x< 10. Then I substitue whatever the adsolute value of (squrae root of 3 < 10. This is where I am stuck. I do not understand how to maniputae (square root 3) to substitue it in for the x in 8 < |(square root 3)|< 10. I think if I get assistance finding out the substitute absolute value I may be able to solve, and find 'g' I have tries to square square (square root x -3) to get 8 <|x-3|< 10then use the 'g' = 0.01/ 13; but that doesn't match with the answer in the back of my text. The answer in the back of my book is 0.058. I am getting 0.00076. Can someone please give me some guidance?
• Sep 29th 2009, 07:51 AM
Plato
First surely you mean $\displaystyle \lim _{x \to 9} \sqrt x = 3$?
Take note that $\displaystyle \left| {\sqrt x - 3} \right| = \frac{{\left| {x - 9} \right|}} {{\left| {\sqrt x + 3} \right|}}$.
So $\displaystyle \left| {x - 9} \right| < 1\, \Rightarrow \,\sqrt 8 < \sqrt x < 10\, \Rightarrow \,\frac{1} {{\sqrt x + 3}} < \frac{1}{{\sqrt 8 + 3}}$
• Sep 29th 2009, 08:59 AM
bosmith
Quote:

Originally Posted by Plato
First surely you mean $\displaystyle \lim _{x \to 9} \sqrt x = 3$?
Take note that $\displaystyle \left| {\sqrt x - 3} \right| = \frac{{\left| {x - 9} \right|}} {{\left| {\sqrt x + 3} \right|}}$.
So $\displaystyle \left| {x - 9} \right| < 1\, \Rightarrow \,\sqrt 8 < \sqrt x < 10\, \Rightarrow \,\frac{1} {{\sqrt x + 3}} < \frac{1}{{\sqrt 8 + 3}}$

Plato, Thank you very much for your instruction. This is great, this is great.