Why does this sequence converge? I have to deal with the following sequence : 

$\lim \limits_{x \to \infty}\sqrt{x+\sqrt{x}} - \sqrt{x}$ 

If I factorize it to $\sqrt{x}(\sqrt{\sqrt{x}+1}-1)$, I would say it diverges since both factors diverge: $\lim \limits_{x \to \infty}\sqrt{x}= \infty  $ and $\lim \limits_{x \to \infty} \sqrt{\sqrt{x}+1} = \infty$  

But if I type it in WolframAlpha, I get $\frac12$ as limit. Can you help me out?
Multiply with the conjugate $\sqrt{x+\sqrt{x}}+\sqrt{x}$ to obtain that $$\begin{align*}\sqrt{x+\sqrt{x}}-\sqrt{x}&amp;=\frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}\overset{(1)}=\frac{x+\sqrt{x}-x}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\\[0.2cm]&amp;=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{\frac{1}{\sqrt{x}}+1}+1\right)}=\frac{\sqrt{\not x}}{\sqrt{\not x}\left(\sqrt{\frac{1}{\sqrt{x}}+1}+1\right)}=\\[0.2cm]&amp;=\frac{1}{\sqrt{\frac{1}{\sqrt{x}}+1}+1} \to \frac{1}{\sqrt{0+1}+1}=\frac12\end{align*}$$ as $x \to \infty$. In (1) we used the identity $$(a-b)(a+b)=a^2-b^2$$ with $a=\sqrt{x+\sqrt{x}}$ and $b=\sqrt{x}$.
A slightly more readable solution results if one sets $y=\sqrt{x}$. Then by general reasoning the limit becomes $\lim_{y\to\infty}\sqrt{y^2+y}-y$.  Then simplify further by setting $z=\frac{1}{y}$, obtaining $\lim_{z\to 0}\sqrt{1/z^2+1/z}-1/z$.
This can be evaluated as follows:
$$
\lim_{z\to 0}\sqrt{1/z^2+1/z}-1/z=\lim_{z\to0}\frac{1}{z}\left(\sqrt{1+z}-1\right)=\lim_{z\to0}\frac{\sqrt{1+z}-1}{z},
$$
which is recognizably the derivative of the square root function at $1$, and therefore the answer is 1/2.