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

Question

I need some help calculating this limit:

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

I know it's equal to 1 but I have no idea how to get there. Can anyone give me a tip? I can't use l'Hopital. Thanks a lot.

Answer 1

EDIT

I guess you need a basic method that preceeds the l'Hospital's rule.

Set $x=a^2,\; a>0.$ The limit rewrites $$ \begin{aligned}\sqrt{a^2+a} - \sqrt{a^2-a}=&\;\left(\sqrt{a^2+a} - \sqrt{a^2-a}\right)\frac{\sqrt{a^2+a}+\sqrt{a^2-a}}{\sqrt{a^2+a}+\sqrt{a^2-a}}\\ =&\;\frac{2a}{\sqrt{a^2+a}+\sqrt{a^2-a}}\\=&\; \frac{2}{\sqrt{1+a^{-1}}+\sqrt{1-a^{-1}}}\to 1\; {\text {as}}\; a \to \infty\end{aligned}$$

My first answer
$$ \begin{aligned}\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}}=&\; \sqrt{\sqrt{x}(\sqrt{x}+1)} - \sqrt{\sqrt{x}(\sqrt{x}-1)}\\=&\; \sqrt[4]{x}\left( \sqrt{\sqrt{x}+1}-\sqrt{\sqrt{x}-1}\right)\\ =&\; \sqrt[4]{x}\left( \sqrt{\sqrt{x}+1}-\sqrt{\sqrt{x}-1}\right)\cdot\frac{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}\\=&\; \sqrt[4]{x}\cdot\frac{2}{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}\\=&\; \frac{2}{\sqrt{1+x^{-1/4}}+\sqrt{1-x^{-1/4}}}\end{aligned}$$ from where the limit is $1.$

Answer 2

HINT

Use that

$$ \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} =\left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)\frac{ \sqrt{x + \sqrt{x}}+ \sqrt{x - \sqrt{x}} }{ \sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}} }=$$

$$=\frac{ x + \sqrt{x}- x + \sqrt{x} }{ \sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}} }$$

Answer 3

Let $x=\left(\frac{t+1}{t-1}\right)^2$ with $t\to 1$.

Then, evaluate the limit

$$\lim_{t\to1}\frac{\sqrt 2 \sqrt{t+1}}{1+\sqrt{t}}$$

Answer 4

By Lagrange's theorem, $a>b>0$ ensures $\sqrt{a}-\sqrt{b} = (a-b)\frac{1}{2\sqrt{c}}$ with $c\in(b,a)$.
If we let $a=x+\sqrt{x}$ and $b=x-\sqrt{x}$ we get $$ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} = \frac{2\sqrt{x}}{2\sqrt{c}},\quad c\in(x-\sqrt{x},x+\sqrt{x})$$ and since $\sqrt{x\pm\sqrt{x}}=\sqrt{x}(1+o(1))$ the outcome is clear.

Answer 5

After you've multiplied by the sum of squares to get $2 \sqrt{x}$ in the numerator, in the denominator you have $\sqrt{x +\sqrt{x}} + \sqrt{x - {\sqrt{x}}}$, so 'pull out' $\sqrt{x}$ to get $\sqrt{x} (\sqrt{1 + \frac{1}{\sqrt{x}}} + \sqrt{1 - \frac{1}{\sqrt{x}}}$. After the cancellation and limit you get 1.

Answer 6

First get rid of $\sqrt{x}$ and of $\infty$ with the substitution $t=1/\sqrt{x}$, that transforms the limit into $$ \lim_{t\to0^+}\left( \sqrt{\frac{1}{t^2}+\frac{1}{t}}-\sqrt{\frac{1}{t^2}-\frac{1}{t}} \right)= \lim_{t\to0^+}\frac{\sqrt{1+t}-\sqrt{1-t}}{t} $$ This is the derivative at $0$ of $f(t)=\sqrt{1+t}-\sqrt{1-t}$; since $$ f'(t)=\frac{1}{2\sqrt{1+t}}+\frac{1}{2\sqrt{1-t}} $$ we have $f'(0)=1$.

Alternatively, multiply by the conjugate: $$ \lim_{t\to0^+}\frac{(1+t)-(1-t)}{t(\sqrt{1+t}+\sqrt{1-t})} $$

Answer 7

I am fond of explicit inequalities that can be confirmed by hand... For real $u \geq 2,$ we find $$ \left( u - \frac{1}{2} - \frac{1}{u} \right)^2 < u^2 - u < \left( u - \frac{1}{2} \right)^2 $$

$$ \left( u + \frac{1}{2} - \frac{1}{8u} \right)^2 < u^2 + u < \left( u + \frac{1}{2} \right)^2 $$

We need $u \geq 2$ because $$ \left( u - \frac{1}{2} - \frac{1}{u} \right)^2 = u^2 - u -\frac{7}{4} + \frac{1}{u} +\frac{1}{u^2} $$ so $u=1$ does not give the inequality we want. $$ $$

$$ u - \frac{1}{2} - \frac{1}{u} < \sqrt{u^2 - u} < u - \frac{1}{2} $$

$$ u + \frac{1}{2} - \frac{1}{8u} < \sqrt{u^2 + u} < u + \frac{1}{2} $$

Take $$ u = \sqrt x $$ so $x \geq 4$

$$ \sqrt x - \frac{1}{2} - \frac{1}{ \sqrt x} < \sqrt{x - \sqrt x} < \sqrt x - \frac{1}{2} $$

$$ \sqrt x + \frac{1}{2} - \frac{1}{8 \sqrt x} < \sqrt{x + \sqrt x} < \sqrt x + \frac{1}{2} $$

Subtract

$$ 1 - \frac{1}{8 \sqrt x} < \sqrt{x + \sqrt x} - \sqrt{x - \sqrt x}< 1 + \frac{1}{ \sqrt x} $$

  x    lower bound         actual             upper bound
  4  0.9375              1.035276180410083  1.5
  5  0.9440983005625052  1.027486296746016  1.447213595499958
  6  0.9489689636920171  1.022520831033128  1.408248290463863
  7  0.9527544408738466  1.019077329344677  1.377964473009227
  8  0.9558058261758408  1.016548303281371  1.353553390593274
  9  0.9583333333333334  1.014611872354577  1.333333333333333
 10  0.9604715292478953  1.01308145723319   1.316227766016838
 11  0.9623110819277796  1.011841408817098  1.301511344577764
 12  0.9639156081756484  1.010816211706107  1.288675134594813
 13  0.9653312377359232  1.009954457590246  1.277350098112615
 14  0.9665923447609469  1.009219933184     1.267261241912424
 15  0.9677251387816048  1.008586390757442  1.258198889747161
 16  0.96875             1.008034339861825  1.25
 17  0.9696830468704584  1.00754900380017   1.242535625036333
 18  0.9705372174505605  1.007118975557603  1.235702260395516
 19  0.9713230332661797  1.006735308900081  1.229415733870562
 20  0.9720491502812526  1.006390888653184  1.223606797749979
 21  0.9727227637205009  1.006079984972172  1.218217890235992
 22  0.9733499104555488  1.005797931788809  1.21320071635561
 23  0.9739356982428656  1.005540890860252  1.208514414057075
 24  0.9744844818460086  1.005305675959844  1.204124145231932
 25  0.975               1.005089620052082  1.2
 26  0.975485483107727   1.004890473669719  1.196116135138184
 27  0.9759437387837656  1.004706326263013  1.192450089729875
 28  0.9763772204369233  1.004535544682177  1.188982236504614
 29  0.9767880827278685  1.004376724590913  1.185695338177052
 30  0.9771782267706181  1.00422865174695   1.182574185835055
 31  0.9775493372466532  1.004090270888218  1.179605302026775
 32  0.9779029130879204  1.003960660536812  1.176776695296637
 33  0.9782402930055377  1.003839012447871  1.174077655955698
 34  0.9785626768571863  1.003724614734089  1.171498585142509