How to evaluate $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$? How can I evaluate  $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?
 A: HINT:
As we are dealing with the limit in real numbers, $x>0\implies x\to+\infty$
Put $x=y^2$
$$\implies \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}$$
$$=\sqrt{y^2+y}-\sqrt{y^2-y}$$
$$=\frac{y^2+y-(y^2-y)}{\sqrt{y^2+y}+\sqrt{y^2-y}} \text{ (Rationalizing the numerator )}$$
$$=\frac2{\sqrt{1+\frac1y}+\sqrt{1-\frac1y}} (\text{Dividing the numerator & the denominator by } y)$$
Now, as $x\to\infty, y\to\infty$

Alternatively, put $x=\frac1{h^2}$ 
$$\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}$$
$$= \frac{\sqrt{1+h}-\sqrt{1-h}}h$$
$$= \frac{1+h-(1-h)}{(\sqrt{1+h}+\sqrt{1-h})h} \text{ (Rationalizing the numerator )}$$
$$=\frac2{\sqrt{1+h}+\sqrt{1-h}}\text{ if }h\ne0$$ 
As $x\to\infty, h\to0\implies h\ne0$
A: I guess someone should mention the Taylor approximation approach:
$$\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right) = \sqrt{x}\left(\sqrt{1+ {1 \over \sqrt{x}}}-\sqrt{1- {1 \over \sqrt{x}}}\right) $$
$$= \sqrt{x}\bigg(\big(1 + {1 \over 2\sqrt{x}} + O({1 \over x})\big) - \big(1 - {1 \over 2\sqrt{x}} + O({1 \over x})\big)\bigg)$$
$$=\sqrt{x}\bigg({1 \over \sqrt{x}} + O({1 \over x})\bigg)$$
$$= 1 + O({1 \over \sqrt{x}})$$
So the limit is $1$.
A: Almost always when you have a limit of $\sqrt{A}-\sqrt{B}$ type. It is good idea to multiply it by $\frac{\sqrt{A}+\sqrt{B}}{\sqrt{A}+\sqrt{B}}$. So you have to do limit of $\frac{A-B}{\sqrt{A}+\sqrt{B}}$
Thus for your limit you get:
$$\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} = \frac{ (x+\sqrt{x} ) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}}=\frac{ 2\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} \rightarrow 1$$ as $x \rightarrow \infty$.
A: Rationalize. Then divide througout by $\sqrt x$. Then substitute limit.
