If $\lim_{x\to\infty} {\{x-f(x)\}}=2$, then find $\lim_{x\to\infty} \frac{\sqrt{x+1}-\sqrt{f(x)}}{\sqrt{x}-\sqrt{f(x)}}$ Question: If $\lim_{x\to\infty} {\{x-f(x)\}}=2$, then find $$\lim_{x\to\infty} \frac{\sqrt{x+1}-\sqrt{f(x)}}{\sqrt{x}-\sqrt{f(x)}}$$
I changed the form to use the condition:
$$\lim_{x\to\infty} \frac{x+1-f(x)}{x-f(x)}\cdot\frac{\sqrt{x}+\sqrt{f(x)}}{\sqrt{x+1}+\sqrt{f(x)}}$$
Even though I can get a value of 
$\lim_{x\to\infty}\frac{x+1-f(x)}{x-f(x)}$ which is 3/2 I guess, I couldn't find the infinite limit of the second fraction. I also thought about using squeeze like
$$\frac{\sqrt{x}}{\sqrt{x+1}+\sqrt{f(x)}}\le\frac{\sqrt{x}+\sqrt{f(x)}}{\sqrt{x+1}+\sqrt{f(x)}}\le\frac{\sqrt{x}+\sqrt{f(x)}}{\sqrt{x+1}}$$
but still did not work well.
I also thought of changing the condition as
$$\lim_{x\to\infty} {x\left(1-\frac{f(x)}{x}\right)}=2$$
Since $\lim_{x\to\infty}x=\infty$, it might follow that $$\lim_{x\to\infty} {f(x)\over x}=1$$
but is this procedure always right? If so, then I think I can get a infinity limit of the second one.
I could predict the answer as just putting $f(x)=x-2$, which in turn we get 3/2. But, as you know, there can exist other possible answers, so I need precise solutions.
Could you please give me some ideas for the question? Thanks.
 A: Since $\displaystyle \lim_{x \to \infty} \left(x-f(x)\right) = 2$, there exist $N>>0$ which satisfies $x > N \implies 1< x-f(x) <3$, i.e. $x-3 < f(x) < x-1$. Then for $x>N$ we have 
$$ \dfrac{\sqrt x + \sqrt{x-3}}{\sqrt{x+1} + \sqrt{x-1} } \le \dfrac{\sqrt x + \sqrt{f(x)}}{\sqrt{x+1} + \sqrt{f(x)} } \le \dfrac{\sqrt x + \sqrt{x-1}}{\sqrt{x+1} + \sqrt{x-3} }$$ and here you can proceed with squeezing. 
A: Your idea to rewrite $x-f(x) = x\left(1-\frac{f(x)}{x}\right)$ is completely correct and also straight forward.
Since $\left(x-f(x)\right)\stackrel{x\to\infty}{\longrightarrow}2$ you have for sufficiently large $x$
$$0\leq x\left(1-\frac{f(x)}{x}\right) <2+\epsilon \Rightarrow  0\leq 1-\frac{f(x)}{x} < \frac{2+\epsilon}{x}\stackrel{x\to\infty}{\longrightarrow} 0$$
Hence, you have $\lim_{x\to \infty}\frac{f(x)}{x}=1$ and you can conclude
\begin{eqnarray*}\frac{\sqrt{x+1}-\sqrt{f(x)}}{\sqrt{x}-\sqrt{f(x)}}
& = & 1+ \frac{\sqrt{x+1}-\sqrt x}{\sqrt x - \sqrt{f(x)}} \\
& = & 1 +\frac{1}{x-f(x)}\cdot\frac{1+\sqrt{\frac{f(x)}{x}}}{1+\sqrt{1+\frac 1x}}\\
& \stackrel{x\to\infty}{\longrightarrow} & 1+\frac 12 = \frac 32
\end{eqnarray*}
