# Does 0/0 and ∞/∞ = 1 ?

[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.

Related Precalculus Mathematics Homework Help News on Phys.org
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.

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

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

cristo
Staff Emeritus
This is driving me crazy. Part of my Calculus work. My graphic calculator says undefined.
Does 0/0 and ∞/∞ = 1 ?
No.
But in my books, the way the solution is, it seems as if they do equal 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.

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.

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

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.

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???

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.

cristo
Staff Emeritus
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 $$\left|\frac{a_{n+1}}{a_n}\right|$$ where here a_n=n/3^n.
After some simplifying we get

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

Hurkyl
Staff Emeritus
Gold Member
Now if we substitute ∞ the first bracket becomes ∞/∞.

arildno
Homework Helper
Gold Member
Dearly Missed
No, the ratio test looks at the quantity:
$$\frac{\frac{n+1}{3^{n+1}}}{\frac{n}{3^{n}}}=\frac{1}{3}\frac{n+1}{n}$$

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

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.

Hurkyl
Staff Emeritus
Gold Member
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....

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

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

I have the solution in front of me. this question was an example, but didn't give explanations.
Put the example down. You need to work through it without the solution.
Now I understand why (n+1)/n = 1
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?

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

cristo
Staff Emeritus
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.

Because the nth power in the numerator is subtracted form the nth power in the denominator.
(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!

HallsofIvy
Homework Helper
Please stop doing that! You have been told repeatedly that you cannot "substitute" $\infty$] for a real number.

The limit of ((n+1)/(n))((3^n)/(3^(n+1))= [(n+1)/n](1/3) is 1/3. Surely you learned, long before you started using L'Hopital's rule, that you do NOT find a limit, as n goes to infinity, by "substuting" [/itex]\infty[/itex] for n.

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" $infty$ for n?
What is [itex]3(\infty)+ 1[/sub]?

What is the limit of n^2/n as n goes to infinity? n/n^2? n/n? This would seem to suggest that you can't "evaluate" an expression at infinity.

i've understood, but misused terms.