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I can't find the limit absolute value function.

 
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Nov22-12, 02:27 PM   #1
 

I can't find the limit absolute value function.


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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Nov22-12, 03:14 PM   #2
 
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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
Nov22-12, 03:48 PM   #3
 
is it just a "plug-in" problem?

(|1-2|-2)/1 => (|-1|-2)/1 => (1-2)/1 =-1?
Nov22-12, 04:53 PM   #4
 
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I can't find the limit absolute value function.


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
Nov22-12, 07:35 PM   #5
 
The value of function is negative(?)
Nov22-12, 09:36 PM   #6
 
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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
Nov22-12, 11:51 PM   #7
 
f(x)=-1
Nov23-12, 07:43 AM   #8
 
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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.
Nov23-12, 01:15 PM   #9
 
thank you HallsofIvy and Ray
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absolute value, calculus, limit
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