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Limit of a sequence, with a real parameter

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  • #1
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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. 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?
 
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  • #2
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Homework Statement



Find a, sucht that:

[tex]

\lim_{ x \to \infty }( a \sqrt{n+2} - \sqrt{n+1} ) ) = \infty(a+1)

[/tex]
Where did you get this problem? We never do arithmetic with ∞.

Also, the limit should probably be as n approaches ∞, not x.
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

 
  • #3
LCKurtz
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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.
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$$
\frac{a\sqrt{n+2}+\sqrt{n+1}}{a\sqrt{n+2}+\sqrt{n+1}}$$
 
  • #4
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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$$
\frac{a\sqrt{n+2}+\sqrt{n+1}}{a\sqrt{n+2}+\sqrt{n+1}}$$

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:

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. :)
 
  • #5
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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.
Two things:
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.
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. :)
This also makes no sense. If a = -1, your limit expression is
$$ \lim_{n \to \infty} [(-1)\sqrt{n + 2} - \sqrt{n + 1}~] = -\lim_{n \to \infty}[\sqrt{n + 2} + \sqrt{n + 1}~]$$
Clearly both square roots get large without bound, so the limit is -∞ when a = -1.
 
  • #6
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1. ∞(a + 1) is not a valid expression. You cannot substitute values of a here and get a result that's meaningful.
This is true, but writing the value of the limit like this, helps to vizualize the cases which needs to be discussed for a. Clearly, you can't substitute values.

2. This site doesn't recognize atex. Use tex or itex in your tags.
I've just discovered that, thank you very much for your advice .

This also makes no sense. If a = -1, your limit expression is
For this, I'm deeply sorry, because I trancsribed the exercise with a mistake. The original exercise was:

[tex]

\lim_{ x \to \infty }( a \sqrt{n+2} + \sqrt{n+1} )

[/tex]

It is correct now, right ?
 
  • #7
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We're still trying to understand what the problem is. Is it this?
Find a such that
$$ \lim_{ n \to \infty }( a \sqrt{n+2} + \sqrt{n+1} ) = 0$$
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.
 
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  • #8
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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.
Yes, this is. I had done what LCKrutz said before I posted, but it didin't help me. And I strongly beleive that my solution is correct, isn't it ?
 
  • #9
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Yes, this is. I had done what LCKrutz said before I posted, but it didin't help me.
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.
And I strongly beleive that my solution is correct, isn't it ?
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.
 
  • #10
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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.
Okay, but can you please explain me why my work isn't valid ? :D . I don't understand that but I would really like to :D .
 
  • #11
Office_Shredder
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As soon as you write down that the limit is equal to [itex] \infty(a+1)[/itex] you have written down a statement that means nothing. If I tell you the limit is [itex] ponies*(a+1)[/itex] and therefore a=-1, how confident would you feel? Writing down infinity like that is the same thing. Here is an example to show the danger:
[tex] \lim_{n\to \infty} n \frac{1}{n}\ ?=?\ \infty*0\ ?=?\ 0 [/tex]

is clearly false. The problem was when I wrote down infinity times zero and thought that I wrote down something meaningful.
 
  • #12
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Sure. Here is what you first posted, with the correction to the limit expression that you gave later.
DorelXD said:
Find a, such that:

[tex]\lim_{ x \to \infty }( a \sqrt{n+2} + \sqrt{n+1} ) ) = \infty(a+1)[/tex]
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:
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.
What you should NOT do is assume any particular value for the limit. Use the hint that LCKurtz gave to find the value of the limit, which will be an expression in a. Then show that the only way for this expression to be zero is when a = -1.

You said you had tried doing what LCKurtz suggested earlier, but that it didn't help. Show us what you did, and we'll steer you in the right direction.
 
  • #13
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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
 
  • #14
LCKurtz
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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
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.
 
  • #15
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Hmmmmm, I still don't get it. :D .
 
  • #16
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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}##.
 
  • #17
LCKurtz
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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.
Hmmmmm, I still don't get it. :D .
Can you choose ##a## so they aren't there?
 
  • #18
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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}##.

Ok, so I get that:

[itex] \sqrt{n} \frac{a^2 + \frac{2a2}{n} - 1 - \frac{1}{n}}{a\sqrt{1 + 2/n} - \sqrt{1+1/n}} [/itex]

What's next ? I'm tempted to write that this tends to infinty times (a+1) .
 

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