Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$ Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$.
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$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
It takes $\infty - \infty $ form when $x=\infty $.
Now,
$$=\lim_{x\to \infty } (\sqrt {x-a} - \sqrt {bx})\times \dfrac {\sqrt {x-a}+\sqrt {bx}}{\sqrt {x-a}+\sqrt {bx}}$$
$$=\lim_{x\to \infty } \dfrac {x-a-bx}{\sqrt {x-a}+\sqrt {bx}}$$

I'll add to Eugen's comment:
If $b=1$ then we have:
$$ \lim_{x\to \infty} \frac{-a}{\sqrt{x-a}+\sqrt{x}} = \frac{-a}{\infty} = 0$$
otherwise,
$$ \lim_{x\to \infty} \frac{(1-b)x-a}{\sqrt{x-a}+\sqrt{x}} =\lim_{x\to \infty} \frac{(1-b)\sqrt{x} - \frac{a}{\sqrt{x}}}{\sqrt{1-\frac{a}{x}}+1} = \lim_{x\to \infty} \frac{(1-b)\sqrt{x}-0}{\sqrt{1-0}+1}=\lim_{x\to\infty} \frac{1-b}{2}\sqrt{x}$$
Meaning that the answer will be $\pm \infty$ dependeing on $\operatorname{sign}(1-b)$
