Limits using basic analysis theorems and logic?

In summary, Joe found an example of a clever way of dealing with limits involving differences like these in the back of the book. He suggests using the binomial theorem to simplify the expression.
  • #1
broegger
257
0
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?
 
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  • #2
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
try doing something to it that involves its conjugate...
 
  • #4
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
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
What is the solution?
 
  • #7
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 irrelevant 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
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
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:

What are limits in mathematics?

Limits in mathematics refer to the value that a function approaches as the input approaches a certain value. It represents the behavior of a function near a specific point.

What is the limit notation?

The limit notation is written as lim(x→a) f(x), where "lim" stands for limit, "x→a" means that the input is approaching a specific value "a", and "f(x)" is the function being evaluated.

What are the basic analysis theorems related to limits?

The basic analysis theorems related to limits are the Squeeze Theorem, the Intermediate Value Theorem, and the Convergence-Divergence Theorem. These theorems provide useful tools for evaluating limits and determining the behavior of a function.

How is logic used in determining limits?

In determining limits, logic is used to analyze the behavior of a function and make inferences based on the given information. Through logical reasoning, it is possible to determine the value of a limit and prove its existence or non-existence.

What is the difference between a one-sided and two-sided limit?

A one-sided limit only considers the behavior of a function from one direction, either from the left or the right of a specific point. A two-sided limit, on the other hand, considers the behavior of a function from both directions and requires that the function approaches the same value from both sides.

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