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Old 07-14-04, 02:03 PM Thread Starter   #1
Frodo Baggins
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Limits


Can someone please check this? I have my doubts, and I need to be solid at this stuff.

Prove lim_(x -> 3-) = 2(x+3)/|x-3| = -2 = L

Where that limit reads as x approaches 3 from the left.
I'll use f(x) = 2(x + 3)/|x-3|

Note,
|f(x) - L | = | 2(x + 3)/|x - 3| + 2 | = |2(x + 3) + 2(x - 3) | / |x - 3| = 2 |(x - 3) + 2|x - 3||/ |x - 3|

But if |x - 3| < 1, then -1 < x - 3 < 1 and x - 3 < 1

So, we are left with,
|f(x) - L| < 2 | 1 + 2|x - 3|| / 1 = 2 + 4|x - 3|

Thus, let e > 0 be arbitrary. Let d = min{1, (e - 2) / 4}. Then if 3 - d < x < 3, then, |x - 3| < (e - 2)/4

|f(x) - L| = 2 |(x - 3) + 2|x - 3||/ |x - 3| < 2 + 4|x - 3| < 2 + (e - 2)/4 = e

QED
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Old 07-14-04, 02:58 PM   #2
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Quote:
Originally Posted by Frodo Baggins
Can someone please check this? I have my doubts, and I need to be solid at this stuff.

Prove lim_(x -> 3-) = 2(x+3)/|x-3| = -2 = L

Where that limit reads as x approaches 3 from the left.
I'll use f(x) = 2(x + 3)/|x-3|

Note,
|f(x) - L | = | 2(x + 3)/|x - 3| + 2 | = |2(x + 3) + 2(x - 3) | / |x - 3| = 2 |(x - 3) + 2|x - 3||/ |x - 3|

But if |x - 3| < 1, then -1 < x - 3 < 1 and x - 3 < 1

So, we are left with,
|f(x) - L| < 2 | 1 + 2|x - 3|| / 1 = 2 + 4|x - 3|

Thus, let e > 0 be arbitrary. Let d = min{1, (e - 2) / 4}. Then if 3 - d < x < 3, then, |x - 3| < (e - 2)/4

|f(x) - L| = 2 |(x - 3) + 2|x - 3||/ |x - 3| < 2 + 4|x - 3| < 2 + (e - 2)/4 = e

QED

It appears to be very boring.

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Old 07-14-04, 05:11 PM   #3
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*Walks into thread*

*Sees very very very very confusing lines and numbers*

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Old 07-14-04, 10:28 PM   #4
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Wantto write it out and post an image of it, I'm not making sence of it, because I have never seen Limits done typed on single lines, so its a bit confusing,


side note.
Want a good textbook for this and related items
Pick up
Essentials of Technical Mathematics with Calculus second Edition
by Richard Paul/M.Leonard Shaevel
ISBN 01-13-289091-7

its an older book it was my First Year Math text book, but has become my book of referance when ever I need a refersher on this stuff

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Old 07-14-04, 10:36 PM   #5
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Quote:
Originally Posted by cw823
It appears to be very boring.
You would appear to be right

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Old 07-14-04, 11:00 PM   #6
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thats why they discovered Differentiation

because limits are very boring, but you need to understand them before you can understand what a Derivative actually is.

then you get to forget how to take limits thats a happy day

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Old 07-15-04, 12:00 AM   #7
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Old 07-15-04, 12:13 AM   #8
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Woah... Way too many parentathes and absolute value signs. My head is officially spinning now. I think the lack of knowing what's being divided by what is really throwing me off. Any way you could put up a copy that's not on one line?


Quote:
Originally Posted by deRusett
thats why they discovered Differentiation

because limits are very boring, but you need to understand them before you can understand what a Derivative actually is.

then you get to forget how to take limits thats a happy day
The day I forgot how to take a limit (ok.. so it hasn't happened yet, but I've been close) was the worst day of my life! Reducing calculus to just sets of rules and regulations denies it of the mathematical beauty that lies underneath. While I think limits are evil (I prefer just plugging infinity/infintessimal straight into the equation ), the idea behind them is interesting and thought provoking.



...though the rules and regulations do make it a LOT easier to blaze through the zillion problems assigned!
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Last edited by JigPu; 07-15-04 at 12:24 AM.
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Old 07-15-04, 12:24 AM   #9
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Old 07-15-04, 12:31 AM   #10
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wow do i got somthing to look forward to in a few years! Algebra 2 next year :P
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Old 07-15-04, 01:01 AM   #11
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i can't even read that thing at all, and my head hurts just by looking at it!!!!

Shard, we are in the same boat. me is very stupid too.

This hurts my head to much, i think i'm going to go look at some Pron now....less thinking required.

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Old 07-15-04, 08:36 AM Thread Starter   #12
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Quote:
Originally Posted by deRusett
Wantto write it out and post an image of it, I'm not making sence of it, because I have never seen Limits done typed on single lines, so its a bit confusing,
Attached Images
File Type: gif tex.gif (10.4 KB, 54 views)
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Old 07-15-04, 08:45 AM   #13
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Old 07-15-04, 12:08 PM   #14
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I don't have a clue, but I give anyone who understands this some mad props.
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Old 07-15-04, 12:08 PM   #15
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from what i remember from earlier this year in calc I, and some of calc II, that seems to be done more in depth than i remember doing epsilon and delta proofs when i had the unpleasant experience of doing them

as i recall, you really don't need to know how to do these that much, just remeber the limits and how to take them and your set

i get the limit to be 1.2 x 10^14th

that is on an ti-86 doing it graphically and via table, and by hand getting something big

i'd say your limit should be either infinity or something along the lines of the above

Last edited by Roboman101; 07-15-04 at 12:19 PM. Reason: got the limit of f(x)
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Old 07-15-04, 12:14 PM   #16
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question1, who said the Lim as X aproches 3 from the left is -2? I get it to be 12 from the equation


questio2
Why did you bother with this line
"But if |x-3| <1, then -1 < x -3 < 1 and x+3 < 7

X+3 can not equal 7 because we ignore the function after X=3

as we aproche 3 from the left we get x=2.99999999999999
which makes |x-3| = 0.0000000000001 ( I didn't count the 9's or zeros just held my finger down )


I must be missing something because unless x is aproching -3 from the left there is no way it can be a negative value

I'd go into more detail but I have 5 minutes to get ready for work, and I don't want to break out a pen and paper

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Old 07-15-04, 02:14 PM   #17
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When did they start using letters in math?
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Old 07-15-04, 02:29 PM   #18
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You are making life hard for yourself on the third line in as deRusett says.

No need to rearrange it to test at 1.

But looking back - is your whole first line given in the question?

I don't think it equals 2...

I did this stuff 2 years ago, now all my calculus is completely different. I'll think about it for a while though.

Quote:
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When did they start using letters in math?
Grade 8 for me...
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Old 07-15-04, 02:32 PM   #19
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I should probably know the answer to this.

I should probably be able to give some input.

I should have actually gone to more than half my maths lectures last year.

Damn.

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Old 07-15-04, 03:07 PM Thread Starter   #20
Frodo Baggins
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Quote:
Originally Posted by deRusett
question1, who said the Lim as X aproches 3 from the left is -2? I get it to be 12 from the equation


questio2
Why did you bother with this line
"But if |x-3| <1, then -1 < x -3 < 1 and x+3 < 7

X+3 can not equal 7 because we ignore the function after X=3

as we aproche 3 from the left we get x=2.99999999999999
which makes |x-3| = 0.0000000000001 ( I didn't count the 9's or zeros just held my finger down )


I must be missing something because unless x is aproching -3 from the left there is no way it can be a negative value

I'd go into more detail but I have 5 minutes to get ready for work, and I don't want to break out a pen and paper
I didn't mean to waste your time. The equation is written wrong. I'll fix some stuff up

Quote:
questio2
Why did you bother with this line
"But if |x-3| <1, then -1 < x -3 < 1 and x+3 < 7

X+3 can not equal 7 because we ignore the function after X=3
It's a bit obsolete now that the equation is wrong. But I never said it was equal to 7.
-1 < x - 3 < 1 => -1 + 6 < x + 3 < 1 + 6 => 5 < x + 3 < 7

We don't worry about the lower bound since -1 < x - 3 < 1 => x + 3 < 7

The reason why was to bound the factor in the expression by a number.

Okay, another try. I've been feeling under the weather, but I hope I didn't commit any serious errors in this one either.
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