Determine positive, real constants $A$, $B$, $C$, for which exist continuous function... Please help me in the problem: Determine positive, real constants $A$, $B$, $C$, for which exist continuous function $f:(0, \infty)\rightarrow\mathbb{R}$, such as:
$$f(x)=\frac{A\sqrt{x}-B}{x^2 -4} \ \ \ \ \ \ \ \ \text{for} \ \ \ \ x>2$$
$$f(x)=\frac{\ln(Cx)}{x-2} \ \ \ \ \ \ \ \ \text{for} \ \ \ \ 0<x<2$$
 A: HINT: You need to choose $A,B$, and $C$ so that 
$$\lim_{x\to 2^{-}}\frac{\ln Cx}{x-2}=\lim_{x\to 2^+}\frac{A\sqrt{x}-B}{x^2-4}\;;$$
the function $f$ can then be given this common limit at $x=2$. Both denominators go to $0$ as $x\to 2$, so the limits cannot exist unless the numerators also go to $0$ as $x\to 2$. We need $\lim_{x\to 2^-}\ln Cx$ to be $0$, and since $\ln u=0$ if and only if $u=1$, this means that we need $C=\frac12$.
Now calculate
$$\lim_{x\to 2^-}\frac{\ln(x/2)}{x-2}\;;$$
l’Hospital’s rule applies. Call the limit $L$.
We also need $\lim_{x\to 2^+}\left(A\sqrt{x}-B\right)=0$, which means that $A\sqrt2-B=0$, or $B=A\sqrt{2}$. Now solve
$$\lim_{x\to 2^+}\frac{A\sqrt{x}-A\sqrt{2}}{x^2-4}=L\;,$$
and you’re nearly done.
Added: If you don’t have l’Hospital’s rule available for the first limit, let $u=x-2$. Then $$\frac{\ln(x/2)}{x-2}=\frac1u\ln\left(1+\frac{u}2\right)=\ln\left(\left(1+\frac{u}2\right)^{1/u}\right)\;,$$ and $u\to 0^-$ as $x\to 2^-$, and you should be able to evaluate 
$$\lim_{u\to 0^-}\ln\left(\left(1+\frac{u}2\right)^{1/u}\right)\;.$$
