limit of a sequence

**iva** · August 28th 2011, 08:44 AM

The answer to this is 1/3 but I don't know why. I'm obviously using the wrong technique but i don't know which one I am supposed to use:

Evaluate lim n-> infinity an

$\sqrt{n^2 + n} - \sqrt{n^2 - 1}$

i divide each term by $n^2$ to get:

$\sqrt{1 + n/1} - \sqrt{1-1/n^2} = \sqrt{1 - 0} - \sqrt{1 - 0} = 0$

I know its wrong but i don't know why

Thank you

**Plato** · August 28th 2011, 08:49 AM

Originally Posted by iva

The answer to this is 1/3 but I don't know why. I'm obviously using the wrong technique but i don't know which one I am supposed to use:

Evaluate lim n-> infinity an

$\sqrt{n^2 + n} - \sqrt{n^2 - 1}$

$\sqrt{n^2 + n} - \sqrt{n^2 - 1}=\frac{n+1}{\sqrt{n^2 + n} + \sqrt{n^2 - 1}}$

**iva** · August 28th 2011, 09:20 AM

Wow thanks i just figured out that identity ( if thats what its called) that i had seen before but have never used. I get that part now.

But now if i divide through by n^2 i get

$(0-0)/1+1 = 0$

which is still wrong..if the answer is 1/3 then should i be dividing through by n rather than n^2? Well i just tried that but then i get

$1/(\sqrt{n+1} + \sqrt{n-1}) = 0$

This doesn't make sense, as i need a 3 in the denominator so i guess one of these terms, i guess the first , should evaluate to 2, and the second to 1. Sorry now i am just taking stabs in the dark

**alexmahone** · August 28th 2011, 09:33 AM

Originally Posted by iva

Wow thanks i just figured out that identity ( if thats what its called) that i had seen before but have never used. I get that part now.

But now if i divide through by n^2 i get

$(0-0)/1+1 = 0$

which is still wrong..if the answer is 1/3 then should i be dividing through by n rather than n^2? Well i just tried that but then i get

$1/(\sqrt{n+1} + \sqrt{n-1}) = 0$

This doesn't make sense, as i need a 3 in the denominator so i guess one of these terms, i guess the first , should evaluate to 2, and the second to 1. Sorry now i am just taking stabs in the dark

$\frac{n+1}{\sqrt{n^2 + n} + \sqrt{n^2 - 1}}=\frac{1+1/n}{\sqrt{1+1/n}+\sqrt{1-1/n^2}}$

**iva** · August 28th 2011, 10:00 AM

you divided the numerator by n but the denominator by n^2, i thought we have to divide by the same thing. OK thanks for that equation, so that gives us 1/(1+1) = 1/2 right?

**alexmahone** · August 28th 2011, 10:02 AM

Originally Posted by iva

you divided the numerator by n but the denominator by n^2, i thought we have to divide by the same thing. OK thanks for that equation, so that gives us 1/(1+1) = 1/2 right?

Actually, I divided the denominator by $\sqrt{n^2}=n$ .

Yes, the limit is 1/2.

**anonimnystefy** · August 28th 2011, 10:03 AM

hi iva

no he divided both by n,but when n goes under the root sign it turns into n^2.

**iva** · August 28th 2011, 10:04 AM

Thanks, i have so much to learn

**iva** · August 28th 2011, 10:26 AM

And you guys found an error in my textbook! their answer was 1/3

Thread: limit of a sequence

LinkBack

Thread Tools

Display

limit of a sequence

Re: limit of a sequence

Re: limit of a sequence

Re: limit of a sequence

Re: limit of a sequence

Re: limit of a sequence

Re: limit of a sequence

Re: limit of a sequence

Re: limit of a sequence

Similar Math Help Forum Discussions

limit of function of a sequence equals the function of the limit of that sequence?

Limit of a sequence

Limit of sequence

limit and sequence

another limit of a sequence

Search Tags