TEN POINTS to whoever gives the first correct, and adequatly explained answer?

Question

f(x) = abs ((x/6) -2)
lim of f(x) as x approches 3 =? (I think it is 3/2)
now find the largest possible delta
if epsilon= .0001 using the delta epsilon method of course.

I get to abs(abs((x/6) -2)- (3/2)) < epsilon, but then I get stuck

ABS= Absolute value

I have worked on this for half an hour, I am doing my own homework

I get stuck because I wind up with 6 abs (x-1/3)-3/2
less than epsilon but that does not get me to x-c less than delta

Answer 1

To prove that the limit of a function f(x) is equal to L
as x approaches a, we have to show that for all E > 0
there exists a number d(normally d(E)) > 0 such that
|f(x) - L| < E whenever |x - a| < d

Note that E is just a number that measures the closeness
of f(x) to L while d measures that of x to a.

Now, we have to prove that the limit of abs((x/6) - 2) as
x approaches 3 is equal to 3/2, as you noted. To do this, we know that
L = 3/2, given E > 0 and d > 0
We have to show that for
|abs((x/6) - 2) - 3/2| < E
x lies in the neighborhood of 3 (i.e very close to 3).

Moving on, we go on simplifying as follows
-E < abs((x/6) - 2) - 3/2 < E
(for we know that when |A| < B, -B < A < B )

Splitting up into two as in
-E < abs((x/6) - 2) - 3/2 and abs((x/6) - 2) - 3/2 < E
3/2 - E < abs((x/6) - 2) and abs((x/6) - 2) < 3/2 + E

Now we split each part into two, because the value in the absolute value bars ((x/6)-2) could be positive or negative...

3/2 - E < ((x/6) - 2) and ((x/6) - 2) < 3/2 + E **OR** -3/2 + E > ((x/6) - 2) and ((x/6) - 2) > -3/2 - E
7/2 - E < (x/6) and (x/6) < 7/2 + E **OR** 1/2 + E > (x/6) and (x/6) > 1/2 - E

combining back, we have
7/2 - E < (x/6) < 7/2 + E **OR** 1/2 + E > (x/6) > 1/2 - E

By simple inspection we can see that for a very small E,
both 7/2 - E and 7/2 + E approach 7/2 = 3.5 (and 0.5 for the other side) and so
3.5- < x/6 < 3.5+ **OR** 0.5- > (x/6) > 0.5+
21- < x < 21+ **OR** 3- > x > 3+

(A- means slightly less that A, A+ means slightly more
than A)

And so we have proved that x lies in the neighbourhood
of 3 as required i.e |x - 3| < d (very small)

Giving E a value of 0.0001 and going back to
3/2 - E < abs((x/6) - 2) < 3/2 + E, we get:
(1.4999) < abs((x/6) - 2) < (1.5001), then...
(1.4999) < ((x/6) - 2) < (1.5001) OR (-1.4999) > ((x/6) - 2) > (-1.5001)
(3.4999) < (x/6) < (3.5001) OR (0.5001) > (x/6) > (0.4999)
(20.9994) < x < (21.0006) OR (3.0006) > x > (2.9994) which agrees with what is above

Keep in mind that I'm rusty on this stuff... it's been about 20 years, LOL!

Answer 2

I don't blame you for having trouble with this. The delta-epsilon definition of limits drove me crazy in school. In this case, epslion refers to the fuction value near the limit, while delta refers to the x value that makes the function approach the limit. I would say that the largest value of delta is the value added to x=3 that gives the result (3/2-epsilon).

Answer 3

first of all which is the delta epsilon method... i'm on curve constuction and i can help you but you must explain to me the delta epsilon method

Answer 4

I hope you fail calculus. Try doing your own homework.

Answer 5

i am kinda looking forward to this answer myself. good luck

Answer 6

Answer- A number

Explain- The answer is a number.

Answer 7

ur a nerd. Go to Harvard.

Answer 8

f(x) = withdrawal from class

Answer 9

do your own homework.

Answer 10

huh... what channel is this?