# How to solve $\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$ Solved

I stuck in this limit, I tried to solve it and gets 1/2 as result. Yet, I was wrong because I forgot a square. Please need help!

$\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$

Note: it's $+\infty$

Thanks in advance

Update: I actually solved it, and this is the way that I wanted to:

$\lim \:_{x\to \:\infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)\:=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x+\sqrt{1+\sqrt{x}}}+\sqrt{x}}\right)=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x^2\left(1+\frac{\sqrt{1+\sqrt{x}}}{x}\right)+\sqrt{x}}}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x\left(\frac{1}{x}+\frac{1}{\sqrt{x}}\right)}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x}\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{x}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1}\right)=\frac{0}{2}=0$

Thanks for your two answers guys. I really appreciate it ! This community is awesome !

• Have you tried multiplying by conjugates to rewrite your expression to contain a difference of squares? – user170231 Sep 28 '15 at 21:45
• @user170231 I'm stuck in this point: $\lim _{x\to \infty \:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}+1}\right)}\right)$ What to do next? – Amine Marzouki Sep 28 '15 at 21:50
• To prevent this question from remaining open, you may add your own solution of the problem as an answer and even accept that. This is much better than just editing a "Solved" into the title ... – Hagen von Eitzen Sep 29 '15 at 17:36
• @HagenvonEitzen I didn't want really to accept one answer, because all of this guys has a correct one, but thanks for your suggestion. I'll choose the first one, just to close this question – Amine Marzouki Oct 2 '15 at 11:13

## 4 Answers

\begin{align}\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x&=\frac{\left(\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x\right)\left(\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x\right)}{\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x}\\ &\le \frac{\left(\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x\right)\left(\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x\right)}{1+\sqrt x}\\ &=\frac{\sqrt{1+\sqrt x}}{1+\sqrt x}\\ &=\frac1{\sqrt{1+\sqrt x}}\\ &\to 0\end{align}

• Could you solve it please in a basic way, we didn't study how to solve limits like what you've just did, so we can't use it – Amine Marzouki Sep 28 '15 at 21:56
• @AmineMarzouki comparison is quite a basic way to show that a limit is zero. It's not clear what you mean by more basic. – Scounged Sep 28 '15 at 22:50
• @Scounged Well, We have different educational systems, so I don't know what do you mean by comparison. We have some limits "tips you could say" but none of them are here in those 2 answers – Amine Marzouki Sep 28 '15 at 22:54
• @AmineMarzouki By comparison, I mean that H.v.E chooses to evaluate a simpler expression, knowing that it's easier to calculate that limit directly than the original, whereafter its possible to draw conclusions about the original limit from the limit calculated instead. What are the limits "tips" you mentioned, and what is your educational level? – Scounged Sep 28 '15 at 23:04
• @Scounged Well, now this year, I have second year of bacalaureat. Tips like: 1. If |f(x)-1|<g(x) and lim g(x)= 0 then lim f(x) = l (as x approach to x0) 2.If f(x)<=g(x) and lim f(x) = +inf then lim g(x) = +inf 3. if g(x)<= f(x) <= h(x) and lim g(x) = lim h(x) = l then lim f(x) = l . I didn't understand you well, what's H.v.E. Could you give me a link to learn those tips? I'll answer you tomorrow I must go to sleep, it's 12:16 AM and I must wake up at 6:30 AM – Amine Marzouki Sep 28 '15 at 23:17

If ${\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} = 0}$ and you are taking a limit of the form $$\lim_{x \rightarrow \infty} \sqrt{x + f(x)} - \sqrt{x}$$ Then you can proceed by writing the limit as $$\lim_{x \rightarrow \infty} \sqrt{x} \bigg(\sqrt{1 + {f(x) \over x}} - 1\bigg)$$ Since $\sqrt{1 + \epsilon} = 1 + {1 \over 2} \epsilon + O(\epsilon^2)$, the limit will be the same as $$\lim_{x \rightarrow \infty} \frac{f(x)}{2\sqrt{x}}$$ In the case at hand, $f(x) = \sqrt{1 + \sqrt{x}}$, which is bounded by for example $\sqrt{\sqrt{x} + \sqrt{x}} = \sqrt 2 x^{1 \over 4}$. Hence the limit will be zero.

For clarity, I will use $y=\sqrt{x}$ The limit becomes

$$\lim_{y\to \infty}( \sqrt{y^2+\sqrt{1+y}}-y)= \lim_{y\to \infty}( \sqrt{y^2+\sqrt{1+y}}-y)\frac{\sqrt{y^2+\sqrt{1+y}}+y}{\sqrt{y^2+\sqrt{1+y}}+y} =\lim_{y\to \infty} \frac{\sqrt{1+y}}{\sqrt{y^2+\sqrt{1+y}}+y} \\ =\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{\sqrt{y^2+y^2\sqrt{1/y^4+1/y^3}}+y}\\ =\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{\sqrt{y^2(1+\sqrt{1/y^4+1/y^3})}+y}\\ =\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{|y|\sqrt{1+\sqrt{1/y^4+1/y^3}}+y}\\ =\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{y\sqrt{1+\sqrt{1/y^4+1/y^3}}+y}\\ =\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{y\left(\sqrt{1+\sqrt{1/y^4+1/y^3}}+1\right)}\\ =\lim_{y\to \infty}\frac{\sqrt{1/y+1}}{\sqrt{y}\left(\sqrt{1+\sqrt{1/y^4+1/y^3}}+1\right)}\\ \to \frac{1}{\infty \cdot 2}=0$$

If we have $f(x)/x^{1/2} \to 0,$ where $f(x) > 0,$ then the MVT gives

$$(x+f(x))^{1/2} - x^{1/2} = 1/(2c_x^{1/2})\cdot f(x) \le f(x)/2x^{1/2} \to 0.$$

Above $c_x \in (x,x+f(x)),$ so we're OK. In this problem $f(x) = (1+x^{1/2})^{1/2},$ so the desired limit is $0.$