Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$

Question

I need to calculate the limit of the function below:

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

I tried multiplying by the conjugate, substituting $x=\frac{1}{t^4}$, and both led to nothing.

Answer 1

Multiply both numerator and denominator by $\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}$

You will get $$\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$$

Divide both numerator and denominator by $\sqrt{x}$

$$\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x\sqrt{x}}}}+1}$$

On finding the limit to infinity, you get

$$\dfrac{\sqrt{1+0}}{\sqrt{1+0}+1} = \dfrac{1}{2}$$

Answer 2

We have by Taylor series

$$ \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\frac{\sqrt{x+\sqrt x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\sim_\infty\frac{\sqrt x}{2\sqrt x}\xrightarrow{x\to\infty}\frac12$$

Answer 3

$$x+\sqrt{x}=(\sqrt{x}+1/2)^2-1/4$$ We can neglect the constant term $-1/4$ so $\lim_{x\to\infty}(\sqrt{x}+1/2)^2-1/4=\lim_{x\to\infty}(\sqrt{x}+1/2)^2$$$\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\lim_{x\to\infty}\sqrt{x+\sqrt{(\sqrt{x}+1/2)^2}}-\sqrt{x}=\lim_{x\to\infty}\sqrt{x+\sqrt{x}+1/2}-\sqrt{x}=\lim_{x\to\infty}\sqrt{(\sqrt{x}+1/2)^2-1/4+1/2}-\sqrt{x}=\lim_{x\to\infty}\sqrt{x}+1/2-\sqrt{x}=1/2$$

Answer 4

I think you're almost right: if I set $\sqrt{x} = t^2$ I got $$ \lim_{t \to \infty} \frac{\sqrt{1+\frac{1}{t^2}}}{\sqrt{1+\frac{1}{t^2}\sqrt{1+\frac{1}{t^2}}} + 1} = \frac{1}{2} $$

Answer 5

Set $\dfrac1x=h^2$ where $h>0$ to get

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

$$=\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}}}}=\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}+\frac1h}}=\sqrt{\frac1{h^2}+\sqrt{\frac{h+1}{h^2}}}$$

$$=\sqrt{\frac1{h^2}+\frac{\sqrt{h+1}}h}=\sqrt{\frac{1+h\sqrt{h+1}}{h^2}}=\frac{\sqrt{1+h\sqrt{h+1}}}h$$

$$\implies\lim_{ x \to \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\lim_{h\to0}\frac{\sqrt{1+h\sqrt{h+1}}-1}h$$

$$=\lim_{h\to0}\frac{1+h\sqrt{h+1}-1}{h(\sqrt{1+h\sqrt{h+1}}+1)}=\frac{\sqrt{0+1}}{\sqrt{1+0\sqrt{0+1}}+1}$$

Answer 6

A useful result for a brute force solution to these sorts of computations is $\sqrt{1+ \theta} = 1+ {1 \over 2} \theta + r (\theta)$, where $\lim_{\theta \to 0} { r(\theta)\over \theta} = 0$

Note that $\sqrt{x+\sqrt{x+ \sqrt{x}}} = \sqrt{x} \sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3} }}$, and so $\sqrt{x+\sqrt{x+ \sqrt{x}}} - \sqrt{x} = \sqrt{x} (\sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3} }} -1) $.

Hence $d(x)=\sqrt{x} (\sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3} }} -1) = \sqrt{x}({1 \over 2}\sqrt{{1 \over x}+\sqrt{1 \over x^3}}+r(\sqrt{{1 \over x}+\sqrt{1 \over x^3}}))$.

Simplifying gives $d(x) = {1 \over 2} \sqrt{1+\sqrt{1 \over x}}+{1 \over 2} {r(\sqrt{{1 \over x}+\sqrt{1 \over x^3}}) \over \sqrt{1 \over x}}$.

Hence $\lim_{x \to \infty} d(x) = {1 \over 2}$.