Proof of limit involving square root

[courtrigrad]

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!

[Tide]

As you have written the expression the limit does not exist. I suspect you meant something else.

[courtrigrad]

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

[Yegor]

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]

lim (sqrt( n+1) - sqrt(n)) * sqrt(n+ 1/2 ) = 1/2
n --> 00
\lim_{n\rightarrow \infty} (\sqrt{n+1} - \sqrt{n}) \sqrt{n+\frac{1}{2}}=
\lim_{n\rightarrow \infty} \frac{((n+1)-n)\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}}=
\lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}}=

Now
2 \sqrt{n+1} > \sqrt{n+1} + \sqrt{n} > 2 \sqrt{n}
so
\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}}
so
\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}}
so
\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}+\sqr t{n}} \leq \frac{1}{2}\lim_{n\rightarrow \infty}\sqrt{\frac{n+\frac{1}{2}}{n}}
\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}}
But now the limits on the RHS and LHS are pretty obviously 1 so we have:
\frac{1}{2} \leq \lim_{n\rightarrow \infty}\frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqr t{n}} \leq \frac{1}{2}
so the limit is \frac{1}{2}

[courtrigrad]

Thanks a lot for the very elegant solution!!!

[maverick280857]

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


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

\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}

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