<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] That is a good idea! Now, intuitively speaking, a square root grows slower than a linear function so you should get "some" infinity. However, it depends on the constants $a$ and $b$ what the result will look like.
E.g. if $a = 0$ and $b=2$, you would get 
\begin{align*}
\lim_{x \rightarrow \infty} \frac{x - 2x}{\sqrt{x} + \sqrt{2x}} &= \lim_{x \rightarrow \infty} \frac{-x}{\sqrt{x} + \sqrt{2x}} \\
&= \lim_{x \rightarrow \infty} \frac{- x}{(1 + \sqrt{2}) \sqrt{x}} = - \lim_{x \rightarrow \infty} \frac{\sqrt{x}}{(1 + \sqrt{2}) } \\
&= - \infty.
\end{align*}
But for other choices of constants $a,b$ you might get $+\infty$. So your answer depends on their values.
In general, a way to proceed would be to use L'Hospital's rules to analyse limits of quotients .
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