Vrftsjtryk

EDIT

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

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

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

By Lagrange's theorem, $a>b>0$ ensures $sqrt{a}-sqrt{b} = (a-b)frac{1}{2sqrt{c}}$ with $cin(b,a)$.

If we let $a=x+sqrt{x}$ and $b=x-sqrt{x}$ we get
$$ sqrt{x+sqrt{x}}-sqrt{x-sqrt{x}} = frac{2sqrt{x}}{2sqrt{c}},quad cin(x-sqrt{x},x+sqrt{x})$$
and since $sqrt{xpmsqrt{x}}=sqrt{x}(1+o(1))$ the outcome is clear.

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

Then, evaluate the limit

$$lim_{tto1}frac{sqrt 2 sqrt{t+1}}{1+sqrt{t}}$$

HINT

Use that

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

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

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

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.

First get rid of $sqrt{x}$ and of $infty$ with the substitution $t=1/sqrt{x}$, that transforms the limit into
$$
lim_{tto0^+}left(
sqrt{frac{1}{t^2}+frac{1}{t}}-sqrt{frac{1}{t^2}-frac{1}{t}}
right)=
lim_{tto0^+}frac{sqrt{1+t}-sqrt{1-t}}{t}
$$
This is the derivative at $0$ of $f(t)=sqrt{1+t}-sqrt{1-t}$; since
$$
f'(t)=frac{1}{2sqrt{1+t}}+frac{1}{2sqrt{1-t}}
$$
we have $f'(0)=1$.

Alternatively, multiply by the conjugate:
$$
lim_{tto0^+}frac{(1+t)-(1-t)}{t(sqrt{1+t}+sqrt{1-t})}
$$