# Thread: limit of a sequence

1. ## limit of a sequence

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

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

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

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

I know its wrong but i don't know why

Thank you

2. ## Re: limit of a sequence

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

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

3. ## Re: limit of a sequence

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

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

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

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

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

4. ## Re: limit of a sequence

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

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

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

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

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

5. ## Re: limit of a sequence

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?

6. ## Re: limit of a sequence

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 $\displaystyle \sqrt{n^2}=n$.

Yes, the limit is 1/2.

7. ## Re: limit of a sequence

hi iva

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

8. ## Re: limit of a sequence

Thanks, i have so much to learn

9. ## Re: limit of a sequence

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