# Thread: Finding limit of function using series expansion

1. ## Finding limit of function using series expansion

Hi,

I'm trying to prove the following limit

$\sqrt{x(x+1)}-x = \frac{1}{2}$ as x approaches infinity.

I'm getting stuck on the expansion of
$x\sqrt{1+1/x}$
I know that it's supposed to be
$x(1+\frac{1}{2x}-\frac{1}{8x^2} +.....)$
but I can't work out the series.
Can anyone help?

2. Originally Posted by thepopasmurf
Hi,

I'm trying to prove the following limit

$\sqrt{x(x+1)}-x = \frac{1}{2}$ as x approaches infinity.

I'm getting stuck on the expansion of
$x\sqrt{1+1/x}$
I know that it's supposed to be
$x(1+\frac{1}{2x}-\frac{1}{8x^2} +.....)$
but I can't work out the series.
Can anyone help?
You don't need a series expansion, but an asymptotic expansion : $\sqrt{1+u}=1+\frac 12 u+o(u)$ when $u\to 0$. Then, substituting, $\sqrt{x(x+1)}-x=x(\sqrt{1+\frac{1}{x}}-1)=x(1+\frac{1}{2x}+o(\frac{1}{x})-1)=\frac{1}{2}+o(1)$, which means that the limit is $\frac{1}{2}$.

3. Oh I see. Thanks. But why does it not work when I try to evaluate the expansion directly?

4. Originally Posted by thepopasmurf
Oh I see. Thanks. But why does it not work when I try to evaluate the expansion directly?
You mean write the series expansion, for $x>1$, $\sqrt{x^2+x}-x=x\left(\sqrt{1+\frac{1}{x}}-1\right)$ $=x\left(1+\frac{1}{2x}+\sum_{k=2}^\infty a_k\frac{1}{x^k} -1\right)= \frac{1}{2}+\sum_{k=2}^\infty \frac{a_k}{x^{k-1}}$ where $a_k=\frac{\frac{1}{2}(\frac{1}{2}-1)\cdots (\frac{1}{2}-k+1)}{k!}$? Then you have to justify that the last series converges to 0 when $x\to+\infty$. There are ways to do that (this is an alternate series so we can get a uniform bound for the remainder of the series by Leibniz theorem), but it is not trivial, and first of all the asymptotic expansion is so much more appropriate (it is dedicated to computation of limits).