<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] An often good trick is to make the substitution $t=1/x$. However, in this case $t=1/\sqrt{x}$ seems better, because we get
$$
\lim_{t\to0^+}\frac{\sqrt{1-at^2}-\sqrt{b}}{t}
$$
The numerator has limit $1-\sqrt{b}$, therefore we see that, if $0\le b<1$ the limit is $\infty$, whereas for $b>1$ the limit is $-\infty$.
If $b=1$, this is the derivative at $0$ of the function $f(t)=\sqrt{1-at^2}$. Since
$$
f'(t)=-\frac{at}{\sqrt{1-at^2}}
$$
we have $f'(0)=0$.
In conclusion
$$
\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})=
\begin{cases}
\infty & 0\le b<1 \\[4px]
0 & b=1 \\[4px]
-\infty & b>1
\end{cases}
$$
</s>