$\lim_{ x \to \infty }\sqrt{x\sqrt{x\sqrt{x}}}-\sqrt{x}$

Question

how to find :

$$\lim_{ x \to \infty }\sqrt{x\sqrt{x\sqrt{x}}}-\sqrt{x}=?$$

my try :

$$\lim_{ x \to \infty }\sqrt{x\sqrt{x\sqrt{x}}}-\sqrt{x}=\lim_{ x \to \infty } x^{7/8}-\sqrt{x}=\infty?$$

Answer 1

$$\lim_{ x \to +\infty }\left(\sqrt{x\sqrt{x\sqrt{x}}}-\sqrt{x}\right)=\lim_{ x \to +\infty } \left(x^{7/8}-x^{1/2}\right)=\lim_{ x \to +\infty } x^{7/8}\left(1-\dfrac{1}{x^{3/8}}\right)=(+\infty)\cdot 1=+\infty.$$

Answer 2

Yes. $$x^{7/8}-\sqrt x=x^{7/8}-x^{1/2}=x^{7/8}\left(1-\frac {1}{x^{3/8}}\right).$$ When $x>2^8$ we have $x^{3/8}>2^3$ and $1-\frac {1}{x^{3/8}}>1-2^{-3}=7/8$ and $$x^{7/8}-\sqrt x> \frac {7}{8} x^{7/8}.$$

Answer 3

Asymptotically, the $-\sqrt{x}$ is irrelevant and the function behaves like $x^{7/8}$. It is thus monotonically increasing and unbounded.

Answer 4

$x^{\frac{7}{8}}-\sqrt{x} = x^{\frac{7}{8}}-x^{\frac{1}{2}} = x^{\frac{1}{2}}(x^{\frac{3}{4}}-1) \sim \mathcal{O}(x^{\frac{5}{4}})$ therefore the limit goes to $+\infty$

Answer 5

We have :

$$\sqrt{x\sqrt{x\sqrt{}x}}-\sqrt{x}=\frac{x(\sqrt{x\sqrt{x}}-1)}{\sqrt{x\sqrt{x\sqrt{}x}}+\sqrt{x}}\underset{+\infty}{\sim}x^{1/8}(\sqrt{x\sqrt{x}}-1)\underset{+\infty}{\sim}x^{1/8}x^{3/4}=x^{7/8}$$

and, in particular :

$$\boxed{\lim_{x\to\infty}\left(\sqrt{x\sqrt{x\sqrt{}x}}-\sqrt{x}\right)=+\infty}$$