Question

I am fairly well versed in Cal 2, but this question bothers me:

lim as n approaches positive infinity for:

(-1)^n * (1 / (3^(n+4) ) )

According to wolfram, this is equal to zero. According to cramster, it's equal to zero.

But according to everything we've learned so far in class, (-1)^(infinity) is equal to "undefined", so undefined * 0 would still equal undefined.

BTW, if anyone's going to say, "Well obviously 1 to any power is always just 1 and only the negative an d positive change the way you've written it", please note that while I agree with that logic, 1^inf is "undefi

Answer 1

ned" at that point. If anyone can explain why it's legal to follow through with multiplying undefined/infinity by 0, please lemme know.

Answer 2

BTW, I *am* perfectly fine with understanding that the complex fraction on the right turns to zero, due to 1 divided by infinity; my only problem lies with why it's legal to multiply the zero against the infinity (undefined) on the left.

Answer 3

[Note: masterfutempaccount was me; I couldn't figure out my password and couldn't get a verification email sent; I finally remembered my pass, so assume I am masterfu from now on :P]

Answer 4

" to multiply the zero against the infinity (undefined) on the left. "

may not be well-defined, but it's certainly not infinity.

it's either 1 or -1

Answer 5

so it doesn't matter... 0*1 = 0 and 0*-1 =0

Answer 6

That's what bothers me, though. If positive one to infinity is considered to be undefined, what makes the fact that it's negative one special? If anything, wouldn't 1^n be "more defined" than (-1)^n?

Answer 7

And I mean I totally agree with your logic that it's -1 or +1, but that seems to disagree with the widely accepted rule that 1^(positive infinity) is indeterminate.

Answer 8

for one thing, it's a limit, so you're not 'pluggin in' infinity, which isn't possible anyway. You're just considering some arbitrarily large number... pick a few and consider them.

Answer 9

Ummm... I've been debating whether I should answer this one or not, only because you seem to have your mind made up. But I can't sit quietly allow you to continue to be misinformed any longer.

First of all you're wrong in saying that\[1^{\infty}\]

is undefined but rather an indeterminate form. Do you understand the difference between the two terms, "undefined" and "indeterminate?"

Answer 10

\[1^{\infty}\]is one the seven indeterminate forms.

Answer 11

Sorry for my late reply. @Algebraic! : I use the term "plugging in" loosely, as we technically do emulate the entering of an infinity. For example, we say that 1/n = 0 as n approaches infinity because we do an imaginary division of 1/infinity. That's what I meant by plugging in infinity.

@calculusfunctions : Englishwise, I do understand the difference, but for whatever odd reason, I seem to swap between the two definitions when doing limit works. To me (if I'm not mistaken), the definitions (when I'm not blindly throwing them about) mean:

indeterminate: "there probably is an answer, but it's humanly/physically impossible to solve, most likely due to there being infinite steps to solve it."

undefined: Impossible to answer because it breaks the logic of math. "1/0", "3 seconds - 10 pounds".


But even then, I guess what I am trying to make sure is:

is limit as n approaches infinity (positive) for 1^n considered to have a limit or to be indeterminate? I mean if this DOES indeed have a limit of 1, then I understand totally why the question I posted here equals zero. Otherwise, I'm still confused as to how (-1)^n is ok, but (+1)^n is divergent/unsolvable for a limit.

Again, sorry for the wait; I was assisting someone else with some simpler math and lost track of replies to my question.

Answer 12

Oh right; almost forgot: Thanks for the help given thus far!

I basically need to confirm whether 1^infinity does end up "limiting" to 1 (logically it does, not sure if it does mathematically). And if so, how does one justify 1^(inf) limits to 1 mathematically (assuming that this turns out to be true). Because if it does end up limiting to 1, this means it totally agrees with my internal logic and will make me very happy. ;)

Answer 13

For example:

Find the limit of the expression (1 + sin4x)ˆ(cot x) as x → 0 from the right.

Solution:
As x → 0 from the right (positive side), (1 + sin4x) → 1 and
cot x → ∞. Hence the expression in the question takes the form 1ˆ∞. Now this expression must first be rewritten in quotient form so that we may apply L' Hospital's rule. First let

y = (1 + sin4x)ˆ(cot x)

Ln y = Ln (1 + sin4x)ˆ(cot x)

Ln y = (cot x) ⋅ Ln (1 + sin4x)

Ln y = [Ln (1 + sin4x)] / (tan x)

Now d(1 + sin4x) / dx = (4cos4x)/(1 + sin4x) and
d(cot x) / dx = sec² x

Thus by L' Hospital's rule again we obtain

lim (Ln y) = lim {[(4cos4x)/(1 + sin4x)] / (sec² x)}

lim (Ln y) = 4

This implies that (Ln y) → 4 as x → 0 from the right. Now

lim [ (1 + sin4x)ˆ(cot x)] as x → 0 from the right

= lim y

= lim eˆ(Ln y)

However (Ln y) → 4. Thus

lim eˆ(Ln y) = eˆ4

∴ the limit of the expression (1 + sin4x)ˆ(cot x) as x → 0 is eˆ4

Answer 14

Ah. Interesting. Thanks! I can see that this shows us that 1^(inf) is indeed indeterminate. So should I assume that if I have 1^n and there's no zero being multiplied into it that I should assume it's divergent? I can see that (defined)*(0) is 0.

In other words, if my problem was:


(1)^n * B, where n approaches infinity,

then I'm assuming if B is anything but 0 that it'd be divergent due to 1^n potentially being anything?

Answer 15

What you have is (-1)ˆn which oscillates, however that is irrelevant because the denominator increases without bound. Thus Wolfram and Cramster were correct! The answer to your problem, my friend, is indeed 0.

Answer 16

Awesome, sounds great, thanks! I'll keep that in mind!

(If I understood correctly, only the fact that we're dividing by infinity on the other fraction leads me to have a valid limit when compared to the 'tricky' value of (-1)^n)

Thanks again!

Answer 17

So I still don't deserve a medal. Hmmm...!

Answer 18

Oh, whoops! Sorry, I tried earlier, but it said I can only give one out (and gave it earlier to algebraicfunctions!). I do have that pseudoaccount still available; I'll see if I can use that one to make a second medal. Heh. Thanks again.

Answer 19

There we go! Thanks again!

Answer 20

Of course! Anytime!