<s>[INST] Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$ 
Evaluate the following limit.
  $$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$

My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-a} + \sqrt {bx}}$$
$$=\lim_{x\to \infty} \dfrac {x-a-bx}{\sqrt {x-a} + \sqrt {bx}}.$$
How do I proceed?
 [/INST] If $b \geq 0$ with $b \neq 1$:
$$ \lim \limits_{x \to +\infty} \sqrt{x-a} - \sqrt{bx} = \lim \limits_{x \to +\infty} \sqrt{x}\Bigg( \sqrt{1 - \frac{a}{x}} - \sqrt{b} \Bigg) = \begin{cases} +\infty & \text{if } 0 \leq b < 1 \\ -\infty & \text{if } b > 1 \end{cases} $$
because, as $x \to +\infty$:
$$ \Bigg( 1 - \frac{a}{x} \Bigg)^{1/2} - \sqrt{b} = 1 - \sqrt{b} - \frac{a}{2x} + o\Big( \frac{1}{x} \Big). $$
As a result:
$$ \lim \limits_{x \to +\infty} \Bigg( 1 - \frac{a}{x} \Bigg)^{1/2} - \sqrt{b} = 1 - \sqrt{b}. $$
If $b=1$: 
$$ \lim \limits_{x \to +\infty} \sqrt{x-a} - \sqrt{bx} = 0. $$
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