
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 $\displaystyle \delta$.
First realize that we have $\displaystyle x_0  \delta_1 = 8.9401$ and $\displaystyle x_0 + \delta_2 = 9.0601$. From that you can solve for $\displaystyle \delta_1\text{ and } \delta_2$ then take $\displaystyle \delta = \min(\delta_1,\delta_2)$.

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. 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 $\displaystyle f(x)L=3x^22x8$ in terms of x2, by substituting (x2)+2 for x in f(x)L and then expanding but keeping the term (x2) together.
This gives $\displaystyle f(x)L=3(x2)^2+10(x2)$ and hence you get
$\displaystyle f(x)L=3(x2)^2+10(x2)=x2\cdot3(x2)+10<\varepsilon$
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 [21,2+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 $\displaystyle \delta := \min(\varepsilon/M,1)$ does the job.

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 $\displaystyle 3x^2 2x 4$ really is 4, then $\displaystyle 3x^2 2x 4$ must approach 4 as x goes to 2 that means that $\displaystyle 3x^2 2x 4 4= 3x^2 2x 8$ must approach 0. And that means that the polynomial $\displaystyle 3x^2 2x 8$ must be equal to 0 when x is equal to 2. and that, in turn, means that x 2 is a factor: $\displaystyle 3x^2 2x 8= (x 2)(3x+ 4)$.
Now, you have $\displaystyle x23x+4< 0.01$ and you want $\displaystyle x 2$ less than something. Of course, as long as $\displaystyle 3x+ 4$ is positive, $\displaystyle x23x+4< .001$ is the same as $\displaystyle x 2< \frac{.001}{3x+ 4}$. 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 $\displaystyle x 2< A< \frac{.001}{3x+ 4}$. Now $\displaystyle A< \frac{.001}{3x+ 4}$ is the same (again, assuming 3x+ 4 is positive) as $\displaystyle 3x+ 4< \frac{.001}{A}$ so this reduces to the problem of finding an upper bound on $\displaystyle 3x+ 4$. Obviously we want x close to 0 so it is sufficient to assume 1< x< 1. When x= 1, $\displaystyle 3x+ 4= 3(1)+ 4= 1$ and when x= 1, 3x+ 4= 3(1)+ 4= 7 so 7 is an upper bound. A= 7, we have $\displaystyle \delta= \frac{.001}{7}$
(Notice the problem said "find a value of $\displaystyle \delta$". There is no one answer. Taking $\displaystyle \delta= \frac{.001}{7}$ 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
 2.99^{2} < sqrt(x) < 3.01^{2}
 8.9401 < x < 9.0601
Now, where do I go from here?
Well, you want $\displaystyle x 9< \delta$ which is the same as $\displaystyle 9 \delta< x< 9+ \delta$ so it looks like you want $\displaystyle \delta$ to satisfy either $\displaystyle 8.9401< 9 \delta$ or $\displaystyle 9+ \delta< 9.0601$. Find the values of $\displaystyle \delta$ that satisfies those and choose the smaller.