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Limits using basic analysis theorems and logic?

  • Thread starter broegger
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  • #1
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Hi,

I need help again. How can I show that

[tex]\sqrt{n^2+2n}-n \rightarrow 1[/tex]​

for [tex]n\rightarrow\infty[/tex] using basic analysis theorems and logic? Any ideas?
 

Answers and Replies

  • #2
Hurkyl
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I bet if you checked your book, you'd find an example of a clever way of dealing with limits involving differences like these...
 
  • #3
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try doing something to it that involves its conjugate...
 
  • #4
CarlB
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Rewrite [tex]\sqrt{n^2+2n}-n [/tex] as


[tex][\sqrt{(n+1)^2 - 1^2} -(n+1)] + 1[/tex]

Thus you need to show that

[tex]\sqrt{(n+1)^2 - 1^2} -(n+1) \rightarrow 0[/tex]

as n to infinity. But if you look at the way it is written, you will see that it looks suspiciously like the difference between the length of two sides of a right triangle as one of the points on the triangle goes to infinity.

Carl
 
  • #5
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Thanks guys. Hurkyl was right; there actually was an example dealing with this in the back of the book. I can't believe I missed it.
 
  • #6
quasar987
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What is the solution?
 
  • #7
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We want to show that [tex]\sqrt{n^2+2n} - (n+1) \rightarrow 0[/tex], so we rewrite the expression:


[tex]\sqrt{n^2+2n} - (n+1) = \frac{\left(\sqrt{n^2+2n} - (n+1)\right)\left(\sqrt{n^2+2n} + (n+1)\right)}{\sqrt{n^2+2n} + (n+1)} = \frac{-1}{\sqrt{n^2+2n} + (n+1)}[/tex]


It is obvious that the final expression tends to 0 as [tex]n\rightarrow\infty[/tex].
 
  • #8
I solved it thus:

(n^2 + 2n)^1/2 = (n^2)^1/2 * (1+2/n)^1/2


therefore this = n(1+2/n)^1/2

now expand the bracket using taylor's theorem to get

(1+2/n)^1/2= 1+(1/n)-(1/n^2)+...


therefore multiplying by n in the above you have n+1-(1/n) as n tends to infinity the (1/n) becomes irrelevent and we are left with n+1. Just subtract the n from the original expression and you obtain the desired result, 1.

Take care, Joe
 
  • #9
HallsofIvy
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Gosh, that looks complicated!

I would have thought using gDogg's suggestion and writing
[tex]\sqrt{n^2+2n}-n= \frac{(\sqrt{n^2+2n}-n)(\sqrt{n^2+2n}+n)}{\sqrt{n^2+2n}+n}[/tex]
[tex]= \frac{2n}{\sqrt{n^2+ 2n}+n}[/tex]

would be much simpler.
 
  • #10
Hurkyl
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Yes... he could have used the binomial theorem!

[tex]
(n^2 + 2n)^{1/2} = (n^2)^{1/2} + \frac{1/2}{1} (n^2)^{-1/2} (2n) + \frac{(1/2) * (-1/2)}{1 * 2} (n^2)^{-3/2} (2n)^2 + \cdots
= n + 1 + O\left(\frac{1}{n}\right)
[/tex]

(since n² > 2n when n grows large)
(Hrm, did he make a mistake in the third term, or did I?)
 
Last edited:

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