From here you want to determine .
First realize that we have and . From that you can solve for then take .
DELTA-EPSILONS
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?
You can develop the polynomial in terms of x-2, by substituting (x-2)+2 for x in f(x)-L and then expanding but keeping the term (x-2) together.
This gives and hence you get
Next you need to find an upper bound for the second factor on the left of the above inequality. To this end you must first limit the range for x (otherwise that second factor would not be bound). For example, you can limit x to the intervall [2-1,2+1]=[1,2]. Now figure out that upper bound M for the second factor on the left for x from this intervall and then divide the entire inequality by that upper bound. You get that does the job.
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.)
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.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?