# Finding the limit

• Jun 24th 2006, 11:44 AM
Nichelle14
Finding the limit
I am to find the limit as n approaches infinity of (sqrt(n+) + sqrt(n-1) - 2sqrt(n).

I separated the last term and then grouped. I did this so I could multiply by a conjugate.

When I simplified I ended up with this:

Lim as n approaches infinity of [(sqrt(n+1) - sqrt(n))^2 - (sqrt(n-1) - sqrt(n))^2]/(sqrt(n+1) - sqrt(n-1))

I don't think I did anything algebracially wrong. But any ways when I try to take the limit, I get a 0 in the denominator.

Can you help?
• Jun 24th 2006, 01:40 PM
CaptainBlack
Quote:

Originally Posted by Nichelle14
I am to find the limit as n approaches infinity of (sqrt(n+) + sqrt(n-1) - 2sqrt(n).

I separated the last term and then grouped. I did this so I could multiply by a conjugate.

When I simplified I ended up with this:

Lim as n approaches infinity of [(sqrt(n+1) - sqrt(n))^2 - (sqrt(n-1) - sqrt(n))^2]/(sqrt(n+1) - sqrt(n-1))

I don't think I did anything algebracially wrong. But any ways when I try to take the limit, I get a 0 in the denominator.

Can you help?

What follows hindges on $\sqrt{x}>x$ when $x<1$, and $\sqrt{x} when $x>1$.

Now:

$
\sqrt{n+1} + \sqrt{n-1} - 2\sqrt{n}=$
$\sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]
$

Now as $1+1/n>1$ when $n>1$:

$
1<\sqrt{1+1/n}<1+1/n
$

Similarly as $1-1/n<1$ when $n>1$:

$
1-1/n<\sqrt{1-1/n}<1
$

So:

$
\sqrt{n}[1+1-1/n-2]<$
$\sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]<\sqrt{n}[1+1/n+1-2]
$

$
-1/\sqrt{n}<$
$\sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]<1/\sqrt{n}
$

Hence:

$
\lim_{n \to \infty}}\ \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]=0
$

so:

$
\lim_{n \to \infty}}\ \sqrt{n+1} + \sqrt{n-1} - 2\sqrt{n}=0
$

RonL