Why does factoring eliminate a hole in the limit?

[Ganesh Ujwal]

\(\displaystyle \lim _{x\rightarrow 5}\frac{x^2-25}{x-5} = \lim_{x\rightarrow 5} (x+5)\)


I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know _why_ this works. I've only been told the methodology of expanding the \(\displaystyle x^2-25\) into \(\displaystyle (x-5)(x+5)\), but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a _bigger picture_ of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the \(\displaystyle (x-5)\), which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure."
I really want to understand this in abstract algebra terms, please elaborate. Thank you very much.

 

[Ishuda]

I'll try for an explanation: The hole doesn't always disappear but when it does it is because one expression is equivalent to another except right at that hole. For example
(x² - 25) / (x - 5) = (x + 5) (x - 5) / (x - 5) = (x + 5)
everywhere except at x = 5. However, if we are very close to 5 and we had infinite place arithmetic (as we do theoretically) the calculation of the first expression [(x² - 25) / (x - 5)] would be very close to the last expression [(x+5)], that is the limit as x approaches 5 of the first expression is the same limit as the limit as x approaches 5 of last expression. Since that latter is a 'nice number', we can define the value at 5 for the first expression and remove the "everywhere except at x = 5" by just putting 10 as the value and not calculating it.

 

[stapel]

\(\displaystyle \lim _{x\rightarrow 5}\frac{x^2-25}{x-5} = \lim_{x\rightarrow 5} (x+5)\)

I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know _why_ this works....

This is where it is helpful to have a clear understanding of the difference between "has a limit value at" and "is continuous at".

You know, from back in algebra, that the function y = (x^2 - 25)/(x - 5) is not continuous at x = 5; its graph is a straight line with a "hole" at x = 5.

However, this exercise is not asking you if the function is continuous at x = 5, or even if it actually exists at x = 5. It is only asking if there is a value for the limit at x = 5. Does the limit exist and, if so, what is it?

The limit value is the value that the expression OUGHT to be, if only (in this case) there weren't that nastiness with division by zero at x = 5. The limit value, in this case, is the value that you'd have with the straight line if you'd filled in that ugly hole at x = 5.

"Has a limit at" means "has a value that it should be". "Is continuous at" means "actually takes on that sensible limit value". This expression, in its original form, cannot be evaluated at x = 5; it is not continuous. But you can get rid of the division-by-zero problem by factoring. Then you can evaluate, find the value that the expression OUGHT to be (absent the "hole"), and therefore find the limit value.

Because the limit is the value that the expression ought to be, absent that hole.

 

[HallsofIvy]

The definition of "limit" is:
"\(\displaystyle \lim_{x\to a} f(x)= L\) if and only if, for any \(\displaystyle \epsilon> 0\) there exist \(\displaystyle \delta> 0\) such that if \(\displaystyle 0< |x- a|< \delta\) then \(\displaystyle |f(x)- L|< \epsilon\)".

Unfortunately many people forget to include that "0< |x-a|" part. Because of that the limit, as x goes to a, is completely independent of the value of f(a). You can change f(a) to be anything at all, or not define f at x= a, without changing the limit.

 

[Steven G]

\(\displaystyle \lim _{x\rightarrow 5}\frac{x^2-25}{x-5} = \lim_{x\rightarrow 5} (x+5)\)


I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know _why_ this works. I've only been told the methodology of expanding the \(\displaystyle x^2-25\) into \(\displaystyle (x-5)(x+5)\), but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a _bigger picture_ of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the \(\displaystyle (x-5)\), which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure."
I really want to understand this in abstract algebra terms, please elaborate. Thank you very much.

Hi,
I would have stated what Ishuda wrote word for word but obviously it was already submitted. I believe that students should see concepts from different viewpoints. I think that it is not a good idea for instructors to push their way down the students throat as it may be best for a student to see it a different way. I also understand that an instructor has limited time and can't show their class each way of thinking about things.
Since we are taking the limit as x approaches 5, then x is never 5. x may be close to 5 but not five.
Let x=12, (12-5)(12+5)/(12-5)=(12+5)...
Letting x=-3, (-3-5)(-3+5)/(-3+5)=(-3+5)....
Letting x=5.0001, (5.0001-5)(5.0001+5)/(5.0001-5)=5.0001+5

Note that above the x-5 factor in the numerator and denominator cancelled out each time leaving the factor (x+5)

Must see that (x-5)(x+5)/(x-5) is a piecewise function. Either (x-5)(x+5)/(x-5) equals (x+5) IF x is not 5 and (x-5)(x+5)/(x-5) is indetermined IF x=5 (since 0/0 is indetermined)