- #1
Flumpster
- 32
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Homework Statement
There isn't a problem statement, I'm just confused about something.
Homework Equations
The Attempt at a Solution
Ok, for a while I didn't understand how cancelling in the following limit works:
(x+5)(x-5)
lim ----------
x--> 5 (x-5)
where you cancel the (x-5)s, because it looked like dividing by zero. Then I (think) I understood correctly, which is that, given f(x), the limit x--> 5 is totally different from f(5). In the latter case you are dividing by zero and in the limit, you are not dividing by zero, the x only approaches 5. Therefore you can cancel the (x-5)'s.
I really hope I've understood at that correctly, if not please explain it to me. Now, here's the bit I REALLY don't understand (:D) :
If I have a limit like:
lim x--> 1, 2x(x-1)
-------
(x/2)-0.5
an online limit calculator tells me the answer is 4. I can also understand how it got to 4,
(take the 2 out, multiply both the numerator and the denominator by 2 and cancel (x-1).) But why can't I cancel the (x-1) and the ((x/2)-0.5)?
The limit about can be written as the (lim x->1 2x)*(lim x ->1 (x-1))/(lim x->1 (x/2)-0.5)
The limits x-> 1 of (x-1) and ((x/2)-0.5) are the same, so why can't I cancel them like the (x-5) in the first example?
Sorry if this is a stupid question, it's just that I'm trying to teach myself this and most material I can find doesn't discuss the ideas behind the math.
Thanks :)