Register to reply

Proof of limit involving square root

by courtrigrad
Tags: involving, limit, proof, root, square
Share this thread:
courtrigrad
#1
Nov11-04, 09:27 PM
P: 1,236
Hello all

I am having trouble proving the limit of the following:

lim sqrt(( n+1) - sqrt(n)) * sqrt(n+ 1/2 ) = 1/2
n --> 00



I tried using the fact the the limit of the first factor as n approaches infinity is 0. Then I tried expressing the first factor as

1 / sqrt(n+1) + sqrt(n) and doing the same thing for the other



factor. However I always get stuck.


Any help would be greatly appreciated!
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
Tide
#2
Nov11-04, 10:58 PM
Sci Advisor
HW Helper
P: 3,144
As you have written the expression the limit does not exist. I suspect you meant something else.
courtrigrad
#3
Nov12-04, 03:17 PM
P: 1,236
lim (sqrt( n+1) - sqrt(n)) * sqrt(n+ 1/2 ) = 1/2
n --> 00

Yegor
#4
Nov12-04, 03:36 PM
P: 147
Proof of limit involving square root

I trhink so:
lim (sqrt( n+1) - sqrt(n)) * sqrt(n+ 1/2 ) =
=lim (sqrt( n+1) - sqrt(n)) *(sqrt( n+1) + sqrt(n)) * sqrt(n+ 1/2 ) /(sqrt( n+1) + sqrt(n)) = lim sqrt(n+ 1/2 )/(sqrt( n+1) + sqrt(n))=1/2
NateTG
#5
Nov12-04, 04:47 PM
Sci Advisor
HW Helper
P: 2,537
Quote Quote by courtrigrad
lim (sqrt( n+1) - sqrt(n)) * sqrt(n+ 1/2 ) = 1/2
n --> 00
[tex]\lim_{n\rightarrow \infty} (\sqrt{n+1} - \sqrt{n}) \sqrt{n+\frac{1}{2}}=[/tex]
[tex]\lim_{n\rightarrow \infty} \frac{((n+1)-n)\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}}=[/tex]
[tex]\lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}}=[/tex]

Now
[tex]2 \sqrt{n+1} > \sqrt{n+1} + \sqrt{n} > 2 \sqrt{n}[/tex]
so
[tex]\frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n+1}} < \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} <
\frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n}}[/tex]
so
[tex]\lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n+1}} \leq \lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} \leq \frac{1}{2}\lim_{n\rightarrow \infty}\frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n}} [/tex]
so
[tex]\lim_{n\rightarrow \infty} \frac{1}{2}\sqrt{\frac{n+\frac{1}{2}}{n+1}} \leq \lim_{n\rightarrow \infty}\frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} \leq \frac{1}{2}\lim_{n\rightarrow \infty}\sqrt{\frac{n+\frac{1}{2}}{n}}[/tex]
[tex]\frac{1}{2}\lim_{n\rightarrow \infty} \sqrt{1 - \frac{\frac{1}{2}}{n+1}} \leq \lim_{n\rightarrow \infty} \frac{1}{2} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} \leq\lim_{n\rightarrow \infty} \frac{1}{2}\sqrt{1 + \frac{\frac{1}{2}}{n}}[/tex]
But now the limits on the RHS and LHS are pretty obviously 1 so we have:
[tex]\frac{1}{2} \leq \lim_{n\rightarrow \infty}\frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} \leq \frac{1}{2}[/tex]
so the limit is [tex]\frac{1}{2}[/tex]
courtrigrad
#6
Nov13-04, 10:20 AM
P: 1,236
Thanks a lot for the very elegant solution!!!
maverick280857
#7
Nov14-04, 03:05 AM
P: 1,780
Quote Quote by NateTG
[tex]\lim_{n\rightarrow \infty} (\sqrt{n+1} - \sqrt{n}) \sqrt{n+\frac{1}{2}}=[/tex]
[tex]\lim_{n\rightarrow \infty} \frac{((n+1)-n)\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}}=[/tex]
[tex]\lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}}=[/tex]
Actually I'd just stop there (I'm not saying Nate's solution is large or anything but here's another way to "see" where the limit is going). I'd then divide the numerator and the denominator by the square root of n to get

[tex]\lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} = \lim_{n\rightarrow \infty} \frac{\sqrt{1+\frac{1}{2n}}}{\sqrt{1+\frac{1}{n}}+1} [/tex]

Taking limits gives (1/2) as the answer. You can recognize the original limit as an indeterminate form and divide by the arbitrarily growing variable n to get to the same thing.

I should mention however, that the sandwiching approach used by NateTG is far more elegant than this "trick" here (which gives you the answer but not an insight).

Cheers
Vivek


Register to reply

Related Discussions
Yes or no, Can you square root a zero Precalculus Mathematics Homework 11
Re: Integral involving square root of e^x Calculus & Beyond Homework 6
What is the square root of i^2 General Math 8
Square root of 3? Fun, Photos & Games 8
Square root General Math 13