Does 0/0 and ∞/∞ = 1 ?

1. May 3, 2008

apples

[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.

2. May 3, 2008

Poop-Loops

What is the homework problem? Is it a limit that you are taking?

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. May 3, 2008

Monocles

Have you learned L'hopital's rule yet? Look it up in your calculus textbook if you haven't, its very simple.

4. May 3, 2008

Poop-Loops

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

5. May 3, 2008

cristo

Staff Emeritus
No.
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.

6. May 3, 2008

apples

I've learn L'hopital's rule, and this isn't related to that.
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. May 3, 2008

sutupidmath

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. May 3, 2008

apples

OK.

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. May 3, 2008

sutupidmath

so i guess your limit is this

$$\lim_{n\to\infty}\left(n+\frac{10}{n}\right)\left(\frac{3^n}{3^{n+1}}\right)$$

right???

10. May 3, 2008

Poop-Loops

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. May 3, 2008

cristo

Staff Emeritus
Ok, so the ration test says to look at $$\left|\frac{a_{n+1}}{a_n}\right|$$ where here a_n=n/3^n.
No. You get
$$\lim_{n\to\infty}\frac{(n+1)}{n}\cdot\frac{3^n}{3^{n+1}} .$$

Simplify this (hint: can you write $3^{n+1}$ in terms of $3^n$ ?

12. May 3, 2008

Hurkyl

Staff Emeritus

13. May 3, 2008

arildno

No, the ratio test looks at the quantity:
$$\frac{\frac{n+1}{3^{n+1}}}{\frac{n}{3^{n}}}=\frac{1}{3}\frac{n+1}{n}$$

14. May 3, 2008

apples

I'm even more confused.
(This isn't precal, it's AP Cal BC)

15. May 3, 2008

apples

i'm sorry i mistyped the corrected function
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))((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.

16. May 3, 2008

Hurkyl

Staff Emeritus
Why do you think that is valid? Reread your limit rules. :tongue:

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. May 3, 2008

apples

OK, I understood what poop-loops is saying. This answers my question:

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. May 3, 2008

cristo

Staff Emeritus
It's not though is it? You missed out crucial parentheses.

Put the example down. You need to work through it without the solution.
This is NOT true, in general. You mean that the limit as n tends to infinity is equal to 1.

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

19. May 3, 2008

apples

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

20. May 3, 2008

cristo

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