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**[SOLVED] Does 0/0 and ∞/∞ = 1 ?**

This is driving me crazy. Part of my Calculus work. My graphic calculator says undefined.

Does 0/0 and ∞/∞ = 1 ?

But in my books, the way the solution is, it seems as if they do equal 1.

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- Thread starter apples
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- #1

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This is driving me crazy. Part of my Calculus work. My graphic calculator says undefined.

Does 0/0 and ∞/∞ = 1 ?

But in my books, the way the solution is, it seems as if they do equal 1.

- #2

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0/0 and ∞/∞ are undefined, but if you take the limit of x/x as x -> ∞ or 0, then it becomes 1, because they are approaching either ∞ or 0 at the same "speed", so they keep canceling each other out.

- #3

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- #4

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Yeah, that's what it's called. It's the general rule for the example I gave. Very useful.

- #5

cristo

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No.This is driving me crazy. Part of my Calculus work. My graphic calculator says undefined.

Does 0/0 and ∞/∞ = 1 ?

Well, you're going to have to give us the exact question, before we can help you. Is there any reason this is in career advice? I'm moving to one of the homework forums.But in my books, the way the solution is, it seems as if they do equal 1.

- #6

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one problem is exactly what poop-loops said.

The limit is 0(right hand limit) though, and the x in the denominator is absolute.

OK, poop-loops, I kind of understand what you're saying.

But here's another problem:

The question is

Determine whether the series:

∞

∑ n/3^n

n=1

converges.

We're supposed to use the ratio test.

After some simplifying we get

lim (n+1/n)(3^n/3^n+1)

n->∞

Now if we substitute ∞ the first bracket becomes ∞/∞.

The solution is 1/3. So in order for the second bracket to become 1/3 this should happen:

(1/3^n-n+1)

And it must be multiplied into 1. So ∞/∞ has to be 1.

I am so confused.

- #7

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Can you rewrite it, it is not clear at which is numerator and which denominator, if there is one at first place!!!

P.S. Just use more parentheses

- #8

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Determine whether the series:

∞

∑ (n)/(3^n)

n=1

converges.

We're supposed to use the ratio test.

After some simplifying we get

lim ((n+10/(n))((3^n)/(3^(n+1)))

n->∞

Now if we substitute ∞ the first bracket becomes ∞/∞.

The solution is 1/3. So in order for the second bracket to become 1/3 this should happen:

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

And it must be multiplied into 1. So ∞/∞ has to be 1.

- #9

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[tex]\lim_{n\to\infty}\left(n+\frac{10}{n}\right)\left(\frac{3^n}{3^{n+1}}\right)[/tex]

right???

- #10

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After some simplifying we get

lim (n+1/n)(3^n/3^n+1)

n->∞

Now if we substitute ∞ the first bracket becomes ∞/∞.

The solution is 1/3. So in order for the second bracket to become 1/3 this should happen:

(1/3^n-n+1)

And it must be multiplied into 1. So ∞/∞ has to be 1.

I am so confused.

You're taking a limit. You're not evaluating the equation AT n = ∞, you're evaluating it as it gets closer and closer to infinity.

So let's take (n+1)/n first. This is your first term. When n = 1, you get 2/1. When n = 100, you get 101/100. When n = 10000, you get 10001/10000. Etc. So when n gets REALLY big, the whole thing gets closer and closer to 1. At n = ∞ you basically ignore the 1 and just end up with n/n.

Now your second term, (3^n)/(3^(n+1)). If I'm not mistaken, you can right off the bat just cancel out 3^n on top and bottom and you end up with 1/(3^1), so there's your 1/3rd.

- #11

cristo

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The question is

Determine whether the series:

∞

∑ n/3^n

n=1

converges.

We're supposed to use the ratio test.

Ok, so the ration test says to look at [tex]\left|\frac{a_{n+1}}{a_n}\right|[/tex] where here a_n=n/3^n.

No. You getAfter some simplifying we get

lim (n+1/n)(3^n/3^n+1)

n->∞

[tex]\lim_{n\to\infty}\frac{(n+1)}{n}\cdot\frac{3^n}{3^{n+1}} .[/tex]

Simplify this (hint: can you write [itex]3^{n+1}[/itex] in terms of [itex]3^n[/itex] ?

- #12

Hurkyl

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Why do you think that's valid? Reread your limit rules.Now if we substitute ∞ the first bracket becomes ∞/∞.

- #13

arildno

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[tex]\frac{\frac{n+1}{3^{n+1}}}{\frac{n}{3^{n}}}=\frac{1}{3}\frac{n+1}{n}[/tex]

- #14

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I'm even more confused.

Let me reread everything.

(This isn't precal, it's AP Cal BC)

Let me reread everything.

(This isn't precal, it's AP Cal BC)

- #15

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my shift key isn't working properly so i put a 0 instead of )

Let me correct it again (sorry)

Determine whether the series:

∞

∑ (n)/(3^n)

n=1

converges.

We're supposed to use the ratio test.

After some simplifying we get

lim ((n+1

n->∞

Now if we substitute ∞ the first bracket becomes ∞/∞.

The solution is 1/3. So in order for the second bracket to become 1/3 this should happen:

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

And it must be multiplied into 1. So ∞/∞ has to be 1.

- #16

Hurkyl

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Surely you've done exercises where, rather than blindly substituting, you painstakingly applied the rules for manipulating limits to justify calculations. Try repeating that exercise here....

- #17

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So let's take (n+1)/n first. This is your first term. When n = 1, you get 2/1. When n = 100, you get 101/100. When n = 10000, you get 10001/10000. Etc. So when n gets REALLY big, the whole thing gets closer and closer to 1. At n = ∞ you basically ignore the 1 and just end up with n/n.

arildno, I have no idea what you're trying to say.

cristo, i didn't understand the first part you wrote, but the second part where you corrected my equation is the equation i wrote.

I have the solution in front of me. this question was an example, but didn't give explanations.

Now I understand why (n+1)/n = 1

- #18

cristo

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It's not though is it? You missed out crucial parentheses.cristo, i didn't understand the first part you wrote, but the second part where you corrected my equation is the equation i wrote.

Put the example down. You need to work through it without the solution.I have the solution in front of me. this question was an example, but didn't give explanations.

This is NOT true, in general. You mean that the limit as n tends to infinity is equal to 1.Now I understand why (n+1)/n = 1

Did you even bother thinking about the hint I gave you?

- #19

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Ok, so the ration test says to look at [tex]\left|\frac{a_{n+1}}{a_n}\right|[/tex] where here a_n=n/3^n.

No. You get

[tex]\lim_{n\to\infty}\frac{(n+1)}{n}\cdot\frac{3^n}{3^{n+1}} .[/tex]

Simplify this (hint: can you write [itex]3^{n+1}[/itex] in terms of [itex]3^n[/itex] ?

yes, i did, and i thought it meant the second bracket would become 1/3

- #20

cristo

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yes, i did, and i thought it meant the second bracket would become 1/3

Sure, but why? You don't have to say, of course, if you are now happy with the solution.

- #21

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(3^n)/(3^(n+1))=1/(3^(n-n+1))=1/3

:D

I already wrote that in one of my previous posts.

Anyway, thank you everybody for the clarification!

- #22

HallsofIvy

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The

Yes, as n goes to infinity, (n+1)/n goes to 1. Now what about (3n+1)/n? What is its limit as n goes to infinity? What do you get if you "substitute" [itex]infty[/itex] for n?

What is [itex]3(\infty)+ 1[/sub]?

- #23

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- #24

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i've understood, but misused terms.

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