Given limit is

$\lim_{x\to\infty}\left(\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}\right)$

Divide numerator and denominator inside the radical by $x$.

$\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}$

$=\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

Now,

$\frac{\sqrt{x+\sqrt{x}}}{x}\to 0$ as $x\to\infty$

Therefore,

$\lim_{x\to\infty}\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

$=\sqrt{\frac{1}{1+0}}$

$=1$