12
$\begingroup$

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.

$\endgroup$
20
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

We have by Taylor series

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

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

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

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

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

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$$x+\sqrt{x}=(\sqrt{x}+1/2)^2-1/4$$ 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$$$\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\lim_{x\to\infty}\sqrt{x+\sqrt{(\sqrt{x}+1/2)^2}}-\sqrt{x}=\lim_{x\to\infty}\sqrt{x+\sqrt{x}+1/2}-\sqrt{x}=\lim_{x\to\infty}\sqrt{(\sqrt{x}+1/2)^2-1/4+1/2}-\sqrt{x}=\lim_{x\to\infty}\sqrt{x}+1/2-\sqrt{x}=1/2$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The end of your second line is missing a $\;\lim\;$ , otherwise that equality is false. $\endgroup$ – Timbuc Dec 27 '14 at 18:45
  • $\begingroup$ @Timbuc Yeah I added limit everywhere,I guess just to be clear $\endgroup$ – kingW3 Dec 27 '14 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.