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?
 A: Multiply with the conjugate $\sqrt{x+\sqrt{x}}+\sqrt{x}$ to obtain that $$\begin{align*}\sqrt{x+\sqrt{x}}-\sqrt{x}&=\frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}\overset{(1)}=\frac{x+\sqrt{x}-x}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\\[0.2cm]&=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{\frac{1}{\sqrt{x}}+1}+1\right)}=\frac{\sqrt{\not x}}{\sqrt{\not x}\left(\sqrt{\frac{1}{\sqrt{x}}+1}+1\right)}=\\[0.2cm]&=\frac{1}{\sqrt{\frac{1}{\sqrt{x}}+1}+1} \to \frac{1}{\sqrt{0+1}+1}=\frac12\end{align*}$$ 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: There's a typo in your factorization, it should be
\begin{equation*}
\sqrt{x+\sqrt{x}}-\sqrt{x} = \sqrt{x} \Big( \sqrt{1+\frac{1}{\sqrt{x}}}-1 \Big)
\end{equation*}
Try multiplying it by its conjugate, i.e.
\begin{equation*}
\sqrt{x+\sqrt{x}}-\sqrt{x} \cdot \frac{\sqrt{x+\sqrt{x}}+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}
\end{equation*}
and expand the numerator.
A: You made a mistake. Note
$$ \sqrt{x+\sqrt{x}}-\sqrt{x}=\sqrt{x}(\sqrt{1+\frac{1}{\sqrt x}}-1) $$
and then rationalize the numerator. You will get the answer.
A: Your factorization is incorrect:
$$\sqrt{x}(\sqrt{\sqrt{x}+1}-1)=\sqrt{x}\sqrt{\sqrt{x}+1} - \sqrt{x}=\sqrt{x\sqrt{x}+x}-\sqrt{x}$$
A: You may write
$$
\sqrt{x+\sqrt{x}} - \sqrt{x}=\frac{(\sqrt{x+\sqrt{x}} - \sqrt{x})(\sqrt{x+\sqrt{x}} + \sqrt{x})}{\sqrt{x+\sqrt{x}} + \sqrt{x}}=\frac{x+\sqrt{x}-x}{\sqrt{x+\sqrt{x}} + \sqrt{x}}\sim \frac{\sqrt{x}}{2\sqrt{x}}=\frac12
$$
as $x$ tends to $+\infty$.
A: If you want to know more than just the limit, you can use the generalized binomial theorem to get an asymptotic series for the function:
$$
\begin{aligned}
(x + x^{1/2})^{1/2}-x^{1/2}&=x^{1/2}((1 + x^{-1/2})^{1/2}-1)\\
&=x^{1/2}\left(\sum_{k=0}^\infty\binom{1/2}{k}x^{-k/2}-1\right)\\
&=\sum_{k=1}^\infty\binom{1/2}{k}x^{(1-k)/2}\\
&=\frac{1}{2}-\frac{1}{8}x^{-1/2}+\frac{1}{16}x^{-1}+O(x^{-3/2})
\end{aligned}
$$
A: 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.
