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?
 A: 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 $b=1$: 
$$ \lim \limits_{x \to +\infty} \sqrt{x-a} - \sqrt{bx} = 0. $$
A: Now, if $b=1$ and $x\rightarrow+\infty$ then the limit is $0$.
If $b\neq1$ then the limit does not exist.
A: 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 .
A: 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
$$
\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})=
\begin{cases}
\infty & 0\le b<1 \\[4px]
0 & b=1 \\[4px]
-\infty & b>1
\end{cases}
$$
A: Alternatively:
$$\lim_\limits{x\to\infty} \sqrt{x-a}-\sqrt{bx}=\lim_\limits{x\to\infty} \sqrt{x}-\sqrt{bx}=(1-\sqrt{b})\lim_\limits{x\to\infty} \sqrt{x}=\begin{cases} 0, \ if \ b=1 \\ \infty, \ if \ b\ne 1. \end{cases}$$
A: Clearly, for $b<1$, $-\infty$ and for $b>1$, $\infty$ (because $\sqrt{1-a/x}-\sqrt b$ tends to the constant $1-\sqrt b$).
Then
$$\sqrt{x-a}-\sqrt{x}=\frac{x-a-x}{\sqrt{x-a}+\sqrt{x}}$$ and the limit is $0$.
