A question about limits

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  • #1
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when we find a limit for an undefined point on a curve , X^2 - 1 / x - 1 at x = 1 for instance
we reshape the equation without actually changing anything to find the limit at this point .
why cant we do that to define the point on the function ?
i mean clearly if we say F(x) = x^2 - 1 / X - 1
we can reshape it by doing F(x) = (x-1) ( X+1 ) / x-1
which is equal to x +1
it only works for the limit , why doesn't it work for defining the function at this point ?
are we only limited to using the shape of the equation that the function is defined by ?
would changing the shape of the function ( the way we express it ) * even though it changes nothing mathematically * change the function ?
 

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  • #2
arildno
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We can't divide by zero. That doesn't bother the limit process, since it looks at the function's behaviour at points in the NEIGHBOURHOOD of the pesky point, not on the pesky point itself. We don't divide by zero in the limit process.

THERE, at the pesky point itself, that expression is simply undefined, and useless as a function value.
 
  • #3
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We can't divide by zero. That doesn't bother the limit process, since it looks at the function's behaviour at points in the NEIGHBOURHOOD of the pesky point, not on the pesky point itself. We don't divide by zero in the limit process.

THERE, at the pesky point itself, that expression is simply undefined, and useless as a function value.
right , so at the point we dont use the expression X^2-1/x-1 when we are discussing a point , but we rather use the expression 1^2-1/1-1 * where 1 is the point* , and thus it is undefined . great
but wait , why dont we even re-express the function so that F(x) = X+1 , now we dont have to divide by zero and it is defined for x = 1 ! ?
 
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  • #4
arildno
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What I've written is basically the difference between f(x)= (x+1)*((x-1)/(x-1)) and g(x)=x+1
Both are also different from h(x)=(x+1)*((x-1)/(x-1))*((x-3)/(x-3))*((x-e)/(x-e))
 
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  • #5
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oh , so changing the expression of the function is not allowed ? great
thanks :D
 
  • #6
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when we find a limit for an undefined point on a curve , X^2 - 1 / x - 1 at x = 1 for instance
Unless you meant x2 - (1/x) - 1, which I'm sure you didn't, use parentheses around the terms in the numerator and the terms in the denominator. Like this: (x2 - 1)/(x - 1)
we reshape the equation without actually changing anything to find the limit at this point .
why cant we do that to define the point on the function ?
i mean clearly if we say F(x) = x^2 - 1 / X - 1
we can reshape it by doing F(x) = (x-1) ( X+1 ) / x-1
which is equal to x +1
You haven't really done anything. Your new "reshaped" version still has a discontinuity at (1, 2).
it only works for the limit , why doesn't it work for defining the function at this point ?
are we only limited to using the shape of the equation that the function is defined by ?
would changing the shape of the function ( the way we express it ) * even though it changes nothing mathematically * change the function ?
The graph of f(x) = (x2 - 1)/(x - 1) has what is known as a removable discontinuity at (1, 2). Most books will show the graph as having a "hole" at (1, 2).

You can extend this function to make it continuous at x = 1 by "plugging the hole."
f(x) = (x2 - 1)/(x - 1), if x ≠ 1
f(x) = 2, if x = 1
 
  • #7
HallsofIvy
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What is true is this: if f(x)= g(x) for all x except a (or just in some "punctured neighborhood" of a: all x close to a but NOT equal to a).
[tex]\frac{x^2- 1}{x- 1}= \frac{(x- 1)(x+ 1)}{x- 1}= x+ 1[/tex]
for all x except x= 1 ("for all x NOT equal to a" because the first is not defined at a but the last is). Therefore, [itex]\lim_{x\to 1}\frac{x^2-1}{x- 1}= \lim_{x\to 1} x+ 1= 2[/itex]
.
As for "defining the function at this point", yes, you can do this- [itex]\frac{x^2- 1}{x- 1}[/itex] has a "removable discontinuity" at x= 1. If we were to define [tex]f(x)= \frac{x^2- 1}{x- 1}[/tex] for x not equal to 1, f(1)= 2, then we have "removed" the discontinuity. But, of course, this is no longer the same function- it is, in fact, exactly the same as the function x+1.
 
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  • #8
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What is true is this: if f(x)= g(x) for all x except a (or just in some "punctured neighborhood" of a: all x close to a but NOT equal to a).
[tex]\frac{x^2- 1}{x- 1}= \frac{(x- 1)(x+ 1)}{x- 1}= x+ 1[/tex]
for all x except x= 1 ("for all x NOT equal to a" because the first is not defined at a but the last is). Therefore, [itex]\lim_{x\to 1}\frac{x^2-1}{x- 1}= \lim_{x\to 1} x+ 1= 2[/itex]
.
As for "defining the function at this point", yes, you can do this- [itex]\frac{x^2- 1}{x- 1}[/itex] has a "removable discontinuity" at x= 1. If we were to define [tex]f(x)= \frac{x^2- 1}{x- 1}[/tex] for x not equal to 1, f(1)= 2, then we have "removed" the discontinuity. But, of course, this is no longer the same function- it is, in fact, exactly the same as the function x+1.
okay great , now just to make sure
if i am given a function in a certain shape F(X) = (x+2)/(X^2 - 4 ) , for example
i am not allowed to change that shape if i want to find the definition of a point on that function right ?
so F(x) = (x+2) / ( X^2 -4 ) is not the same as H(x) = 1 / ( X - 2 ) , correct ?
 
  • #9
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okay great , now just to make sure
if i am given a function in a certain shape F(X) = (x+2)/(X^2 - 4 ) , for example
i am not allowed to change that shape if i want to find the definition of a point on that function right ?
so F(x) = (x+2) / ( X^2 -4 ) is not the same as H(x) = 1 / ( X - 2 ) , correct ?
They are exactly the same except at one point. The graph of F has a hole at (-2, -1/4). The graph of H is defined at that point.

Both graphs have the line x = 2 as a vertical asymptote.
 
  • #10
HallsofIvy
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okay great , now just to make sure
if i am given a function in a certain shape F(X) = (x+2)/(X^2 - 4 ) , for example
i am not allowed to change that shape if i want to find the definition of a point on that function right ?
"Not allowed"? Who's going to stop you?
You are "allowed" to do whatever you want with function- but they won't necessarily be the same when you are finished!

so F(x) = (x+2) / ( X^2 -4 ) is not the same as H(x) = 1 / ( X - 2 ) , correct ?
Yes, that's true. (I wish you wouldn't mix "x" and "X" like that- they are not the same symbol and cannot be assumed to mean the same thing.) F(x) is defined everywhere except at x= 2 and x= -2 while H(x) is defined everywhere except at x= 2.

It is true that "F(x)= H(x) for all x except x= -2". That detail is, unfortunately, often passed over in Calculus classes because, as I said before, since they are the same for all x except x= 2, they have the same limit at x= -2 and that is what we are primarily concerned with.
 
  • #11
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"Not allowed"? Who's going to stop you?
You are "allowed" to do whatever you want with function- but they won't necessarily be the same when you are finished!


Yes, that's true. (I wish you wouldn't mix "x" and "X" like that- they are not the same symbol and cannot be assumed to mean the same thing.) F(x) is defined everywhere except at x= 2 and x= -2 while H(x) is defined everywhere except at x= 2.

It is true that "F(x)= H(x) for all x except x= -2". That detail is, unfortunately, often passed over in Calculus classes because, as I said before, since they are the same for all x except x= 2, they have the same limit at x= -2 and that is what we are primarily concerned with.
okay , thanks alot
and by not allowed i just mean a change would occur if i do so !
 
  • #12
arildno
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Another thing to remember is that whenever we talk about a function, it is strictly speaking necessary to specify the actual DOMAIN of x's values we choose to work with, and the number set (for example the reals) in which the RANGE of the function values lies.
-----------------------------------
Usually, we implicitly think that the domain of x's is what we call the "maximal domain" within the number set the x's belongs to.
In our two cases, we see that the maximal domain of f(x)=(x^2-1)/(x-1) is all reals except x=1, while the maximal domain of g(x)=x+1 is ALL reals.

However, why should we always care about the MAXIMAL domain?

Why can't we regard these functions as being functions on the domain 5<=x<=17?

Note that on THIS domain choice (rather than the choice of maximal domains), we have full equivalence between f(x) and g(x); they are the SAME function, when we look at them as functions from (5,17) into the reals
 

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