Limit of $h(x)$ as $x \to \infty$

Question

Let $h(x) = \sqrt{x + \sqrt{x}} - \sqrt{x}$ , find $\lim_{x \to \infty } h(x)$ . I've tried substitution , multiplying by conjugate and l'hospital's rule but didn't work .

Answer 1

Multiplying and dividing by the "conjugate" $\sqrt{x + \sqrt{x}} + \sqrt{x}$ is a good strategy!

Note that for $x>0$, $$ \sqrt{x + \sqrt{x}} - \sqrt{x}=\frac{ (x + \sqrt{x}) - x}{ \sqrt{x + \sqrt{x}} + \sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}}} + \sqrt{x}}=\frac{ 1}{ \sqrt{1 + \frac{1}{\sqrt{x}}} + 1}.$$ Can you take it from here?

Answer 2

Note that $\sqrt{x+\sqrt{x}} - \sqrt{x} = \frac{x+\sqrt{x}-x}{\sqrt{x+\sqrt{x}}+\sqrt{x}} = \frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}} = \frac{1}{\sqrt{1+\sqrt{\frac{1}{x}}}+1}$, and that $\frac{1}{x}$ is tending towards $0$.

This should lead you to the answer.