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August 13th, 2015, 04:33 PM   #1
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Question Justification for cancelling terms in limits?

I am confused about the algebraic process of finding a limit. Let us take $\displaystyle \frac{x^2 -1}{x - 1}$. In trying to find $\displaystyle \lim_{x\rightarrow 1}\frac{x^2 -1}{x - 1}$ we do the following:

$\displaystyle \lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}$

$\displaystyle \lim_{x\rightarrow 1}x+1$

2

But what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that $\displaystyle \lim_{x\rightarrow 1}x+1$ leads to the correct answer to the original problem if $\displaystyle x + 2$ is different function, with a different domain, than $\displaystyle \frac{x^2 -1}{x - 1}$?
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August 13th, 2015, 04:40 PM   #2
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The domain hasn't changed.
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August 13th, 2015, 05:07 PM   #3
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Originally Posted by skipjack View Post
The domain hasn't changed.
The domain for the first function is all real numbers except 1, as that would cause an undefined value 0 / 0.

The domain of the second is all real numbers.

How are the domains not different?
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August 13th, 2015, 05:28 PM   #4
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The reason is that $x = 1$ is a single discontinuous point. Taking a limit does not require that the function is defined at $x = 1$, but it does need to be defined in intervals around 1.
Since $f(x) = \dfrac{x^2-1}{x-1}$ and $g(x) = x + 1$ have the same values near $x = 1$, they have the same limit at $x = 1$.
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August 13th, 2015, 06:05 PM   #5
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Mr Davis 97 chose domains that are different, but didn't need to do so.

The limit would remain as 2 even if the function had been defined as being equal to 8 when x = 1. In that case, the function's value at x = 1 would still be 8 after cancellation of x - 1.
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August 14th, 2015, 04:02 AM   #6
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A fundamental law for limits, often not given sufficient attention (I have used one textbook that did not even mention it) is

If f and g are such that f(x)= g(x) everywhere in some neighborhood of a except at x= a, then $\displaystyle \lim_{x\to a} f(x)= \lim_{x\to a} g(x)$.

That is easily proved from the definition of "limit".

In this case, $\displaystyle \frac{x^2- 1}{x- 1}= x+ 1$ for all x except x= 1. Therefore, $\displaystyle \lim_{x\to 1} \frac{x^2- 1}{x- 1}= \lim_{x\to 1} x+ 1= 2$.

Last edited by skipjack; August 14th, 2015 at 05:50 AM.
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August 14th, 2015, 07:10 AM   #7
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The concept of a limit is, in laymans terms, the value that is looks like a function is going to get to if we follow the line.

Thus, the function $$f(x) = \begin{cases}{x^2 -1 \over x-1} & x \ne 1 \\ 2 & x = 1 \end{cases}$$
looks like it is going towards $f(1)=2$, but so does $$g(x) = \begin{cases}{x^2 -1 \over x-1} & x \ne 1 \\ 8 & x = 1 \end{cases}$$ and so does $$h(x) = \begin{cases}x+1 & x \ne 1 \\ 8 & x = 1 \end{cases}$$
The reason is that, when $x \ne 1$ we have $${x^2 - 1 \over x-1}= (x+1)\cdot {x-1 \over x-1} = (x+1)\cdot 1 = x+1$$
The reason we can cancel the $x-1$ terms in the numerator and denominator is that $x \ne 1$ which means that $x-1 \ne 0$ and we know that ${a \over a} = 1$ for all $a \ne 0$.

As Country Boy says. Whenever $x \ne 1$ the functions all have the same value. This means that whenever we are heading towards $x=1$ it always looks like we are going to get to the same place.

Perhaps another way to look at it is to imagine playing a little game. Starting at $x=0$ (and moving to the right), draw the graph of the function $f(x)$. You can stop wherever you like before you reach $x=1$. Now, looking only at what you have drawn, guess what the value will be at $x=1$. If you aren't sure, go closer. Go as close as you like to $x=1$, but stop before you get there and make your guess. Repeat with $g(x)$ and $h(x)$.
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August 14th, 2015, 12:58 PM   #8
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I said before 'That is easily proved from the definition of "limit".'

Again there is a detail in the definition of "limit" that is often over looked: $\displaystyle lim_{x\to a} f(x)= L$ means "Given any $\displaystyle \epsilon> 0$, there exist $\displaystyle \delta> 0$ such that if $\displaystyle 0< |x- a|< \delta$ then $\displaystyle |f(x)- L|< \epsilon$."

It is that "$\displaystyle 0< |x- a|$" that is "often over looked" (in fact, I almost didn't write it here). What happens at x= a is irrelevant

Last edited by greg1313; August 14th, 2015 at 01:31 PM.
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