Limit[x^0.5 ((x + 4)^0.5 - x^0.5), x -> ∞]
1. does not exist
2. exists and is equal to 0
3. exists and is equal to 1/2
4. exists and is equal to 2
(The answer is given as D. But how to proove it? )
With simple manipulations you obtain...
$\displaystyle \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x})= \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x}) \cdot \frac{\sqrt{x+4}+\sqrt{x}}{\sqrt{x+4}+\sqrt{x}} = $
$\displaystyle = \frac{4\cdot \sqrt{x}}{\sqrt{x+4}+\sqrt{x}}= \frac{4}{1+\sqrt{\frac{x+4}{x}}}$
But...
$\displaystyle \lim_{x \rightarrow \infty} \sqrt{\frac{x+4}{x}}=1$
... so that...
$\displaystyle \lim_{x \rightarrow \infty} \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x})= 2$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$