# find limit of sequence

• March 17th 2009, 08:10 PM
twilightstr
find limit of sequence
An= square root (n+ square root n + 1) - n find limit of sequence.

my work: lim as x goes to infinite x(1+ (x+1)^ 1/2 / (x))^ (1/2) - squareroot x

i think the limit is infinite minus infinite but how do i change it into the indeterminate form.
• March 17th 2009, 08:18 PM
GaloisTheory1
Quote:

Originally Posted by twilightstr
An= square root (n+ square root n + 1) - n find limit of sequence.

my work: lim as x goes to infinite x(1+ (x+1)^ 1/2 / (x))^ (1/2) - squareroot x

i think the limit is infinite minus infinite but how do i change it into the indeterminate form.

$\lim_{n \rightarrow \infty}(n+ \sqrt{n + 1}) \cdot \frac{n- \sqrt{n + 1}}{n- \sqrt{n + 1}}$

$\lim_{n \rightarrow \infty}\frac{n^2- (n + 1)}{n- \sqrt{n + 1}}$

can you take this limit?
• March 17th 2009, 09:11 PM
twilightstr
the problem is square root of n + squareroot of n+1)) - square root of n
everything before square root n is within a square root
• March 17th 2009, 09:37 PM
Chris L T521
Quote:

Originally Posted by twilightstr
the problem is square root of n + squareroot of n+1)) - square root of n
everything before square root n is within a square root

Thus, $\lim\left(\sqrt{n+\sqrt{n+1}}-\sqrt{n}\right)\frac{\sqrt{n+\sqrt{n+1}}+\sqrt{n}} {\sqrt{n+\sqrt{n+1}}+\sqrt{n}}=\lim\frac{\sqrt{n+1 }}{\sqrt{n+\sqrt{n+1}}+\sqrt{n}}$ $=\lim\frac{1}{\sqrt{\frac{n}{n+1}+\frac{1}{\sqrt{n +1}}}+\sqrt{\frac{n}{n+1}}}=\boxed{\frac{1}{2}}$

Does this make sense?
• March 19th 2009, 07:47 PM
twilightstr
what happened after the second to last step?
• March 20th 2009, 04:01 AM
stapel
Quote:

Originally Posted by twilightstr
what happened after the second to last step?

Divide through, top and bottom, by the square root of n + 1. (Wink)