Help calculating $\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$ 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.
 A: Let $x=\left(\frac{t+1}{t-1}\right)^2$ with $t\to 1$.
Then, evaluate the limit
$$\lim_{t\to1}\frac{\sqrt 2 \sqrt{t+1}}{1+\sqrt{t}}$$
A: 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.
A: HINT
Use that
$$ \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} =\left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)\frac{ \sqrt{x + \sqrt{x}}+ \sqrt{x - \sqrt{x}} }{ \sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}} }=$$
$$=\frac{ x + \sqrt{x}- x + \sqrt{x} }{ \sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}} }$$
A: 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 $1.$
A: 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.
A: 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:
$$
\lim_{t\to0^+}\frac{(1+t)-(1-t)}{t(\sqrt{1+t}+\sqrt{1-t})}
$$
