Find the value of : $\lim_{x\to\infty} \sqrt{x+\sqrt{x}}-\sqrt{x}$ I tried to multiply by the conjugate:
$\displaystyle\lim_{x\to\infty} \frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{x-x+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}$
I don't even know if my rewriting has helped at all. How would I go about doing this?

$\displaystyle \lim_{x\to\infty}\left(\sqrt{x+\sqrt x}-\sqrt x\right) = \lim_{x\to\infty}\sqrt{x}\left\{\left(1+\frac{1}{\sqrt{x}}\right)^{\frac{1}{2}}-1\right\}$
Now expand Using Binomialy
$\displaystyle = \lim_{x\to\infty}\sqrt{x}.\left\{\left(1+\frac{1}{2.\sqrt{x}}+\frac{1}{2}.\left(\frac{1}{2}-1\right).\frac{1}{(\sqrt{x})^2}+.........\right)-1\right\} = \lim_{x\to\infty}\sqrt{x}\left\{\frac{1}{2.\sqrt{x}}+\frac{1}{2}.\left(\frac{1}{2}-1\right).\frac{1}{(\sqrt{x})^2}+.........\right\}$
$\displaystyle =\frac{1}{2}$
