# Epsilon-Delta Proofs :(

• Sep 5th 2010, 06:39 PM
Shaidester
Epsilon-Delta Proofs :(
DELTA-EPSILONS
1) What is the limit of f(x) = 3x2 - 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
• 2.992 < sqrt(x) < 3.012
• 8.9401 < x < 9.0601

Now, where do I go from here?
• Sep 6th 2010, 05:49 AM
lvleph
From here you want to determine $\delta$.

First realize that we have $x_0 - \delta_1 = 8.9401$ and $x_0 + \delta_2 = 9.0601$. From that you can solve for $\delta_1\text{ and } \delta_2$ then take $\delta = \min(\delta_1,\delta_2)$.
• Sep 6th 2010, 10:39 AM
Failure
1. a)
Quote:

Originally Posted by Shaidester
DELTA-EPSILONS
1) What is the limit of f(x) = 3x2 - 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.

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?

You can develop the polynomial $f(x)-L=3x^2-2x-8$ 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 $f(x)-L=3(x-2)^2+10(x-2)$ and hence you get

$|f(x)-L|=|3(x-2)^2+10(x-2)|=|x-2|\cdot|3(x-2)+10|<\varepsilon$
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 $\delta := \min(\varepsilon/M,1)$ does the job.
• Sep 6th 2010, 11:29 AM
HallsofIvy
Quote:

Originally Posted by Shaidester
DELTA-EPSILONS
1) What is the limit of f(x) = 3x2 - 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 $3x^2- 2x- 4$ really is 4, then $3x^2- 2x- 4$ must approach 4 as x goes to 2- that means that $3x^2- 2x- 4- 4= 3x^2- 2x- 8$ must approach 0. And that means that the polynomial $3x^2- 2x- 8$ must be equal to 0 when x is equal to 2. and that, in turn, means that x- 2 is a factor: $3x^2- 2x- 8= (x- 2)(3x+ 4)$.
Now, you have $|x-2||3x+4|< 0.01$ and you want $|x- 2|$ less than something. Of course, as long as $3x+ 4$ is positive, $|x-2||3x+4|< .001$ is the same as $|x- 2|< \frac{.001}{3x+ 4}$. 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 $|x- 2|< A< \frac{.001}{3x+ 4}$. Now $A< \frac{.001}{3x+ 4}$ is the same (again, assuming 3x+ 4 is positive) as $3x+ 4< \frac{.001}{A}$ so this reduces to the problem of finding an upper bound on $3x+ 4$. Obviously we want x close to 0 so it is sufficient to assume -1< x< 1. When x= -1, $3x+ 4= 3(-1)+ 4= 1$ and when x= 1, 3x+ 4= 3(1)+ 4= 7 so 7 is an upper bound. A= 7, we have $\delta= \frac{.001}{7}$

(Notice the problem said "find a value of $\delta$". There is no one answer. Taking $\delta= \frac{.001}{7}$ 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
• 2.992 < sqrt(x) < 3.012
• 8.9401 < x < 9.0601

Now, where do I go from here?
Well, you want $|x- 9|< \delta$ which is the same as $9- \delta< x< 9+ \delta$ so it looks like you want $\delta$ to satisfy either $8.9401< 9- \delta$ or $9+ \delta< 9.0601$. Find the values of $\delta$ that satisfies those and choose the smaller.