<s>[INST] Help calculating $\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$ I need some help calculating this limit:
$$\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$$
I know it's equal to 1 but I have no idea how to get there. Can anyone give me a tip? I can't use l'Hopital. Thanks a lot.
 [/INST] EDIT
I guess you need a basic method that preceeds the l'Hospital's rule.
Set $x=a^2,\; a>0.$ The limit rewrites 
$$ \begin{aligned}\sqrt{a^2+a} - \sqrt{a^2-a}=&\;\left(\sqrt{a^2+a} - \sqrt{a^2-a}\right)\frac{\sqrt{a^2+a}+\sqrt{a^2-a}}{\sqrt{a^2+a}+\sqrt{a^2-a}}\\
=&\;\frac{2a}{\sqrt{a^2+a}+\sqrt{a^2-a}}\\=&\;
\frac{2}{\sqrt{1+a^{-1}}+\sqrt{1-a^{-1}}}\to 1\; {\text {as}}\; a \to \infty\end{aligned}$$
My first answer
$$ \begin{aligned}\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}}=&\;
\sqrt{\sqrt{x}(\sqrt{x}+1)} - \sqrt{\sqrt{x}(\sqrt{x}-1)}\\=&\;
\sqrt[4]{x}\left( \sqrt{\sqrt{x}+1}-\sqrt{\sqrt{x}-1}\right)\\
=&\; \sqrt[4]{x}\left( \sqrt{\sqrt{x}+1}-\sqrt{\sqrt{x}-1}\right)\cdot\frac{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}\\=&\;
\sqrt[4]{x}\cdot\frac{2}{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}\\=&\;
\frac{2}{\sqrt{1+x^{-1/4}}+\sqrt{1-x^{-1/4}}}\end{aligned}$$ from where the limit is $1.$
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