<s>[INST] 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?
 [/INST] 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 $b=\sqrt{x}$.
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