one-sided limit: Find lim [x -> 3-] [abs(x-3) / (x - 3)]
- Thread starter KingAce
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skeeter
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Re: ONE SIDED LIMITS QUESTION!
the limit as x approaches 3 from the left is -1.
as x approaches 3 from the left, the denominator becomes a small negative value. the numerator approaches the same value, but the absolute value makes it positive ... result is a quotient = -1.
the limit as x approaches 3 from the left is -1.
as x approaches 3 from the left, the denominator becomes a small negative value. the numerator approaches the same value, but the absolute value makes it positive ... result is a quotient = -1.
Re: ONE SIDED LIMITS QUESTION!
Any time you have an abs(something), write out the definition, as in:
abs(x - 3) = { x-3 when (x-3)>=0, and -(x-3) when (x-3)<0.}
Now use it in your expression:
abs(x - 3)
----------- =
(x - 3)
{ x-3 when (x-3)>=0, and -(x-3) when (x-3)<0}
--------------------------------------------------------- =
(x - 3)
Now x -->3- means x -> 3 from the left, therefore x < 3, therefore x - 3 < 0.
So { x-3 when (x-3)>=0, and -(x-3) when (x-3)<0} simplifies to -(x - 3) and your expression becomes:
-(x - 3)
---------- = - 1 and lim - 1 = -1.
(x - 3)
No worrying about small numbers,etc. Just very strict application of the def.
KingAce said:
Any time you have an abs(something), write out the definition, as in:
abs(x - 3) = { x-3 when (x-3)>=0, and -(x-3) when (x-3)<0.}
Now use it in your expression:
abs(x - 3)
----------- =
(x - 3)
{ x-3 when (x-3)>=0, and -(x-3) when (x-3)<0}
--------------------------------------------------------- =
(x - 3)
Now x -->3- means x -> 3 from the left, therefore x < 3, therefore x - 3 < 0.
So { x-3 when (x-3)>=0, and -(x-3) when (x-3)<0} simplifies to -(x - 3) and your expression becomes:
-(x - 3)
---------- = - 1 and lim - 1 = -1.
(x - 3)
No worrying about small numbers,etc. Just very strict application of the def.
Helldobler
New member
- Joined
- Dec 24, 2012
- Messages
- 1
Misapplied Def.
IF you are going to turn the absolute value into a piecewise then you need to set the expression inside the abs value to zero to find out how the graph has been shifted.
In this case, x-3=0--->x=3.
Therefore your piecewise wise would be: {x-3, x>=3
{-(x-3), x<3
Now, since we are looking for the limit as x approaches 3 from the negative sided, we can certainly use the second portion of the piecewise, namely -(x-3), x<3 (since we are looking for values before 3). Naturally, we can deduce that -(x-3)/x-3 would be -1. This can be confirmed by graphing the original function.
IF you are going to turn the absolute value into a piecewise then you need to set the expression inside the abs value to zero to find out how the graph has been shifted.
In this case, x-3=0--->x=3.
Therefore your piecewise wise would be: {x-3, x>=3
{-(x-3), x<3
Now, since we are looking for the limit as x approaches 3 from the negative sided, we can certainly use the second portion of the piecewise, namely -(x-3), x<3 (since we are looking for values before 3). Naturally, we can deduce that -(x-3)/x-3 would be -1. This can be confirmed by graphing the original function.