Someone in the tutoring center at my school asked me to do a limit problem for their calc 1 class, and I came up with the following solution. I would like someone to come up with an alternate solution, I don't really like mine, since I applied a theorem I believe to be true but don't recall ever seeing it done. I ended up with the right answer though, but I don't know if that was a coincidence.
Here's the "theorem" I used:
Theorem: If then . Provided and
Then ..........since we are not dealing with infinite limits, the limit of the quotient is equal to the quotient of the limits (that's true right? )
Thus we have
Now here's the question, and my solution.
(Yeah, I know, you guys thought I'd have a much harder problem for you, well no, i'm only smart enough to tackle easy problems)
Now recall that .
Applying the theorem I attempted to prove above we have that as well (This is the part I'm not sure about, I've never this limit defined as that before)
Can anyone come up with a better and/or shorter method? (L'Hopital's is not allowed).
is not a limit that shows up anywhere in my textbook, but it is obvious now that i see it. That was nice! I almost doubt there will be a shorter way to do it.
Now i'm a guy who likes to see several ways of doing the same problem. Does anyone else have another solution? remember, no L'Hopital Rule
That being said, i would like you guys to comment on my proof, how was it? would the theorem be true if or ? Of course we would define "division by infinity" as 0 and "division by 0" as infinity, we're dealing with limits so we can do that, since L may never actually attain the value 0 and no number can actually attain infinity
Haha, i completely agree, and i could fit them in my proof if i wanted to, but i didn't think it was completely necessary and i was already babbling a lot. but your right, if there is one thing i have learned in the proof class i took last semester it's that there's always an epsilon and he's always greater than zero, otherwise, the universe will explode. even if you don't see him, he's there in the background, as he is in my proof. delta is okay, he shows up every now and then, but epsilon is always there!A serious proof should involve epsilon-delta.
No need to use . Because there is an equivalent sequencial definiton*.
Let be defined on an open interval containing such that on . And . Then that means for any sequence converging to we have . So then by the simple property of sequence limits. Hence, by the sequencial definition .
Remark: Jhevon did a minor mistake. He said which is good, but the above proof shows that we need that fact for some open interval containing except possible at .
*)Let be defined on we define iff there exists a non-empty open interval such that for all converging to we have converge to .