Where did you get this problem? We never do arithmetic with ∞.DorelXD said:Homework Statement
Find a, sucht that:
[tex]
\lim_{ x \to \infty }( a \sqrt{n+2} - \sqrt{n+1} ) ) = \infty(a+1)
[/tex]
DorelXD said:Now, I want this sequence to have the limit 0. The first impule is to say that a+1 = 0 and hence a = -1. But if I do this I get [tex] \infty 0 [/tex] which can't be determined. The paradox is that a = -1 works, because if a = -1 then indeed the sequence tends to 0. So, what am I missing? Can you help me, please?
Homework Statement
Homework Equations
The Attempt at a Solution
DorelXD said:Homework Statement
Find a, sucht that:
[tex]
\lim_{ x \to \infty }( a \sqrt{n+2} - \sqrt{n+1} ) ) = \infty(a+1)
[/tex]
Now, I want this sequence to have the limit 0.
Thank you for your answer! But, I figured it out myself. The sequence tends to [atex] \infty (a+1) [/atex], with a a real parameter. We want the sequence to tend to 0. Now, it's obvious that we have three main cases:LCKurtz said:So do you really mean the problem is to find ##a## such that$$
\lim_{n\to\infty}(a\sqrt{n+2}-\sqrt{n+1})=0\text{ ?}$$If so, try multiplying by$$
\frac{a\sqrt{n+2}+\sqrt{n+1}}{a\sqrt{n+2}+\sqrt{n+1}}$$
Two things:DorelXD said:Thank you for your answer! But, I figured it out myself. The sequence tends to [atex] \infty (a+1) [/atex], with a a real parameter.
This also makes no sense. If a = -1, your limit expression isDorelXD said:We want the sequence to tend to 0. Now, it's obvious that we have three main cases:
1) a>0 which means that our sequence tends to infinity
2) a = -1 we get infinity times 0 which is not defined.
3) a < -1 we get -infinity
So, we find a necessary condition, a = -1, and we check to see if in this case our sequence tends to 0 . By pluging a and solving the limit we get indeed that our limit tends to 0. :)
1. ∞(a + 1) is not a valid expression. You cannot substitute values of a here and get a result that's meaningful.
2. This site doesn't recognize atex. Use tex or itex in your tags.
This also makes no sense. If a = -1, your limit expression is
Find a such that
$$ \lim_{ n \to \infty }( a \sqrt{n+2} + \sqrt{n+1} ) = 0$$
Mark44 said:We're still trying to understand what the problem is. Is it this?
If so, follow the advice given by LCKurtz in post #3, with one slight change -- the sign between the radicals should be - in both numerator and denominator.
It worked for me. I took the limit using LCKurtz's tip and got an expression that involved a (and that does NOT include ∞). Then it's easy to show that the only value of a that yields 0 in that expression is a = -1.DorelXD said:Yes, this is. I had done what LCKrutz said before I posted, but it didin't help me.
DorelXD said:And I strongly believe that my solution is correct, isn't it ?
Mark44 said:It worked for me. I took the limit using LCKurtz's tip and got an expression that involved a (and that does NOT include ∞). Then it's easy to show that the only value of a that yields 0 in that expression is a = -1. Your solution of a = -1? Yes, I agree here, but the value is almost beside the point. The work that you showed before was not valid, which tends to cast a dark shadow over any result you got.
First off, you CANNOT use ∞ in an expression and then do arithmetic on it. It makes zero sense to say (or write) that the value of a limit is ∞ * (a + 1).DorelXD said:Find a, such that:
[tex]\lim_{ x \to \infty }( a \sqrt{n+2} + \sqrt{n+1} ) ) = \infty(a+1)[/tex]
DorelXD said:Now, I want this sequence to have the limit 0. The first impule is to say that a+1 = 0 and hence a = -1. But if I do this I get [tex] \infty 0 [/tex] which can't be determined.
The paradox is that a = -1 works, because if a = -1 then indeed the sequence tends to 0.
DorelXD said:Thank you both for your answers. I get it now. Well, I multiplied with the conjugate expresion (I'm used with that trick, it was the first thing I tried) and it led me to: [tex] \frac{ a^2n+2a^2 -n -1 }{a\sqrt{n+2}-\sqrt{n+1}} [/tex]
From here, I couldn't continue. :D
LCKurtz said:Hint: As long as there are ##n##'s in the numerator vs. only ##\sqrt n##'s in the denominator, the fraction isn't going to go to zero.
DorelXD said:Hmmmmm, I still don't get it. :D .
Mark44 said:I factored n out of all four terms in the numerator, and ##\sqrt{n}## out of the two terms in the denominator.
The idea is that ##\sqrt{n + 1} = \sqrt{n}\sqrt{1 + 1/n}##.
The limit of a sequence with a real parameter is the value that the terms of the sequence approach as the parameter approaches a specific value, typically denoted as n → a or x → c.
The limit of a sequence with a real parameter can be calculated by plugging in the specific value of the parameter into the sequence and observing the resulting values. Alternatively, one can use methods such as the squeeze theorem, the ratio test, or the root test to calculate the limit.
A finite limit of a sequence with a real parameter means that the terms of the sequence approach a specific value as the parameter approaches a certain value. An infinite limit means that the terms of the sequence approach either positive or negative infinity as the parameter approaches a certain value.
Yes, the limit of a sequence with a real parameter can exist even if the sequence does not converge. This can happen when the terms of the sequence oscillate between two values as the parameter approaches a certain value, resulting in a limit that does not exist.
The behavior of the terms of a sequence can greatly influence the limit of a sequence with a real parameter. For example, if the terms of the sequence become increasingly larger or smaller as the parameter approaches a certain value, the limit may approach positive or negative infinity. If the terms of the sequence approach a specific value or alternate between two values as the parameter approaches a certain value, the limit will approach that specific value or not exist, respectively.