Question 1122093: Consider the following limit. lim x→3 (x^2 + 9)
Find the limit L.
Well I know that the limit L is equal to 18, but then it asks me the following:
(a) Find δ > 0 such that |f(x) − L| < 0.01 whenever 0 < |x − c| < δ. (Round your answer to five decimal places.)
(b) Find δ > 0 such that |f(x) − L| < 0.005 whenever 0 < |x − c| < δ. (Round your answer to five decimal places.)
What is this trying to tell me to do? I am a very lost and would like to better understand. Thank you!
Found 2 solutions by Theo, 1234554321gijoe:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! limit of x^2 + 9 is equal to 18 as x approaches 2.
you want to find a delta > 0 such that |x^2 + 9 - 18| < .01 when |x - 3| < delta
x^2 + 9 - 18 simplifies to x^2 - 9
therefore you want to find a delta > 0 such that |x^2 - 9| < .01 when |x - 3| < delta
|x^2 = 9| < .01 becomes -.01 < x^2 = 9 < .01 which becomes 8.99 < x^2 < 9.01
solve for x to get = sqrt(8.99) < x < sqrt(9.01)
subtract 3 from all sides to get sqrt(8.99) - 3 < x - 3 < sqrt(9.01) - 3
your delta will be either |sqrt(8.99) - 3| or it will be|sqrt(9.01) - 3|.
|sqrt(8.99) - 3| is equal to .0016671299
|sqrt(9.01) - 3| is equal to .001666204.
when |sqrt(8.99) - 3| < .0016671299, you get 2.99833287 < x < 3.00166713
when x = 2.99833287, you get |x^2 - 9| = .01 which is smaller than or equal to .01
when x = 3.00166713, you get |x^2 - 9| = .0100055586 which is not smaller than or equal to .01.
so, delta = .9915561288 will not work.
when |sqrt(9.01) - 3| is equal to .001666204, you get 2.998333796 < x < 3.001666204.
when x = 2.998333796, you get |x^2 - 9| = .0099944475 which is smaller than or equal to .01
when x = 3.001666204, you get |x^2 - 9| = .01 which is smaller than or equal to .01
so, delta = .001666204 will work.
as long as |x - 3| < .001666204, |x^2 - 9| will be smaller than .01.
your delta is .001666284 which rounds to .00167
use the same procedure for |x^2 - 9| < .005
|x^2 - 9| < .005 means -.005 < x^2 - 9 < .005
add 9 to all sides to get 8.995 < x^2 < 9.005
take the square root of all sides to get sqrt(8.995) < x < sqrt(9.005)
subtract 3 from all sides to get sqrt(8.995) - 3 < x - 3 < sqrt(9.005) - 3
your delta will be either |sqrt(8.995) - 3| or it will be |sqrt(9.005) - 3|
that says your delta will be either .000833449106 or it will be .008332176247
when delta = .000833449106, 2.999166551 < x < 3.000833449
when x = 2.999166551, |x^2 - 9| = .005 which is smaller than or equal to .005
when x = 3.000833449, |x^2 - 9| = .0050013893 which is not smaller than or equal to .005
so, delta = .000833449106 will not work.
when delta = .008332176247, 2.999166782 < x < 3.000833218
when x = 2.999166782, |x^2 - 9| = .0049986115 which is smaller than .005
when x = 3.000833218, |x^2 - 9| = .005 which is smaller than or equal to .005
so, delta = .008332176247 will work.
as long as |x - 3| < .008332176247, |x^2 - 9| will be < .005
your delta is .008332176247 which rounds to .00833
this can also be solved graphically.
the following videos discuss the concept.
not all explanations are simple to follow, but hopefully you will get the idea from these videos, which i chose because they were simpler to follow than others, at least for me.
https://www.youtube.com/watch?v=ftAuCXNAvtE
https://www.youtube.com/watch?v=NaufUAAKdgk
https://www.youtube.com/watch?v=MAMPkdzsAas
https://www.youtube.com/watch?v=gE6n3Da0U4k
in some cases, the deltas will be the same, in which case it's a no brainer what the delta needs to be.
in other cases, as in your problem, the deltas will be different which means you have to determine which of the deltas is the one that will satisfy the requirements of |f(x) - L| < epsilon.
Answer by 1234554321gijoe(1)  (Show Source):
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