
EpsilonDelta Proofs :(
DELTAEPSILONS
1) What is the limit of f(x) = 3x^{2}  2x  4 as x approaches 2?
Obviously it is 4, now to 1a)
1a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001.
2) What is the limit of f(x) = sqrt(x) as x approaches 9?
Obviously it is 3, now to 2a)
2a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001.
Now, I actually started both #1 and #2. My work, respectively:
1a) Let epsilon = 0.01
 f(x)  L < epsilon
 3x^2  2x  4  4 < 0.01
 0.01 < 3x^2  2x  8 < 0.01
Now, where do I go from here?
2a) Let epsilon = 0.01
 f(x)  L < epsilon
 sqrt(x)  3 < 0.01
 0.01 < sqrt(x)  3 < 0.01
 2.99 < sqrt(x) < 3.01
 2.99^{2} < sqrt(x) < 3.01^{2}
 8.9401 < x < 9.0601
Now, where do I go from here?

From here you want to determine .
First realize that we have and . From that you can solve for then take .

1. a)

Quote:
Originally Posted by
Shaidester DELTAEPSILONS 1) What is the limit of f(x) = 3x^{2}  2x  4 as x approaches 2? Obviously it is 4, now to 1a) 1a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001. 2) What is the limit of f(x) = sqrt(x) as x approaches 9? Obviously it is 3, now to 2a) 2a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001. Now, I actually started both #1 and #2. My work, respectively: 1a) Let epsilon = 0.01
 f(x)  L < epsilon
 3x^2  2x  4  4 < 0.01
 0.01 < 3x^2  2x  8 < 0.01
Now, where do I go from here?
Note that if the limit of really is 4, then must approach 4 as x goes to 2 that means that must approach 0. And that means that the polynomial must be equal to 0 when x is equal to 2. and that, in turn, means that x 2 is a factor: .
Now, you have and you want less than something. Of course, as long as is positive, is the same as . But that won't do because you need a NUMBER on the right, not a function of x.
It would be sufficient to find a number "A" such that . Now is the same (again, assuming 3x+ 4 is positive) as so this reduces to the problem of finding an upper bound on . Obviously we want x close to 0 so it is sufficient to assume 1< x< 1. When x= 1, and when x= 1, 3x+ 4= 3(1)+ 4= 7 so 7 is an upper bound. A= 7, we have
(Notice the problem said "find a value of ". There is no one answer. Taking works but so would any smaller number and possibly some larger numbers.)
Quote:
2a) Let epsilon = 0.01
 f(x)  L < epsilon
 sqrt(x)  3 < 0.01
 0.01 < sqrt(x)  3 < 0.01
 2.99 < sqrt(x) < 3.01
 2.99^{2} < sqrt(x) < 3.01^{2}
 8.9401 < x < 9.0601
Now, where do I go from here?
Well, you want which is the same as so it looks like you want to satisfy either or . Find the values of that satisfies those and choose the smaller.