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I can't find the limit absolute value function.

by Vola
Tags: absolute value, calculus, limit
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Vola
#1
Nov22-12, 02:27 PM
P: 14
Hi everybody,

I can't find the limit of (abs(x-2)-2)/x as x-->1.
I know it's (-1) but I don't see how you get to it.
If I take (x-2)>0 I get L=-3,(x-2)<0 I get L=-1.
However according to the graph two sided limit does exist and it's (-1).

Please help
Thanks
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Ray Vickson
#2
Nov22-12, 03:14 PM
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P: 4,959
Quote Quote by Vola View Post
Hi everybody,

I can't find the limit of (abs(x-2)-2)/x as x-->1.
I know it's (-1) but I don't see how you get to it.
If I take (x-2)>0 I get L=-3,(x-2)<0 I get L=-1.
However according to the graph two sided limit does exist and it's (-1).

Please help
What happens when x is near 1?

RGV
Vola
#3
Nov22-12, 03:48 PM
P: 14
is it just a "plug-in" problem?

(|1-2|-2)/1 => (|-1|-2)/1 => (1-2)/1 =-1?

Ray Vickson
#4
Nov22-12, 04:53 PM
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P: 4,959
I can't find the limit absolute value function.

Quote Quote by Vola View Post
is it just a "plug-in" problem?

(|1-2|-2)/1 => (|-1|-2)/1 => (1-2)/1 =-1?
What is your answer to the question I asked you?

RGV
Vola
#5
Nov22-12, 07:35 PM
P: 14
The value of function is negative(?)
Ray Vickson
#6
Nov22-12, 09:36 PM
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P: 4,959
Quote Quote by Vola View Post
The value of function is negative(?)
Let me ask it one last time: if f(x) = (|x-2|-2)/x, what would be the simplified formula for f(x) when x is near 1? Don't guess---calculate!

RGV
Vola
#7
Nov22-12, 11:51 PM
P: 14
f(x)=-1
HallsofIvy
#8
Nov23-12, 07:43 AM
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PF Gold
P: 39,348
Good! For all x less than 2, f(x)= -1 so the limit at any x less than two is -1.

If I take (x-2)>0 I get L=-3,(x-2)<0 I get L=-1.
But you don't "take (x- 2)> 0"! That would be x> 2 and you want the limit at x= 1. You can't "go to 1" if keep x> 2.
Vola
#9
Nov23-12, 01:15 PM
P: 14
thank you HallsofIvy and Ray


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