# Homework Help: Limits using basic analysis theorems and logic?

1. Aug 20, 2005

### broegger

Hi,

I need help again. How can I show that

$$\sqrt{n^2+2n}-n \rightarrow 1$$​

for $$n\rightarrow\infty$$ using basic analysis theorems and logic? Any ideas?

2. Aug 20, 2005

### Hurkyl

Staff Emeritus
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. Aug 20, 2005

### GDogg

try doing something to it that involves its conjugate...

4. Aug 20, 2005

### CarlB

Rewrite $$\sqrt{n^2+2n}-n$$ as

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

Thus you need to show that

$$\sqrt{(n+1)^2 - 1^2} -(n+1) \rightarrow 0$$

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. Aug 21, 2005

### broegger

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. Aug 21, 2005

### quasar987

What is the solution?

7. Aug 21, 2005

### broegger

We want to show that $$\sqrt{n^2+2n} - (n+1) \rightarrow 0$$, so we rewrite the expression:

$$\sqrt{n^2+2n} - (n+1) = \frac{\left(\sqrt{n^2+2n} - (n+1)\right)\left(\sqrt{n^2+2n} + (n+1)\right)}{\sqrt{n^2+2n} + (n+1)} = \frac{-1}{\sqrt{n^2+2n} + (n+1)}$$

It is obvious that the final expression tends to 0 as $$n\rightarrow\infty$$.

8. Aug 22, 2005

### josephcollins

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. Aug 23, 2005

### HallsofIvy

Gosh, that looks complicated!

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

would be much simpler.

10. Aug 23, 2005

### Hurkyl

Staff Emeritus
Yes... he could have used the binomial theorem!

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

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

Last edited: Aug 23, 2005