# 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:

$$f(x)=\frac{A\sqrt{x}-B}{x^2 -4} \ \ \ \ \ \ \ \ \text{for} \ \ \ \ x>2$$ $$f(x)=\frac{\ln(Cx)}{x-2} \ \ \ \ \ \ \ \ \text{for} \ \ \ \ 0<x<2$$

-
Do you know what you need to do? Do they not define the function at $x=2$? –  Git Gud Jan 12 '13 at 18:34
what have you tried? where did you get stuck? –  user8 Jan 12 '13 at 18:34
@GitGud No. I can define this function as I like in x=2 –  Steve Jan 12 '13 at 18:36
Hint. Mean Value Theorems for $C^1$ functions or integrable functions $f$. –  MathOverview Jan 12 '13 at 18:36
@n.c. I've tried to reduce ln(Cx) to x-2 somehow, to make one sided limit of 2 to be a real number. –  Steve Jan 12 '13 at 18:36

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

$$\lim_{u\to 0^-}\ln\left(\left(1+\frac{u}2\right)^{1/u}\right)\;.$$

-
Very much thanks!. 1. How to solve $\lim_{x\to 2^-}\frac{\ln(x/2)}{x-2}$ without l'Hospital, and how to solve the last limit equation? –  Steve Jan 12 '13 at 18:56
@Steve: For the last one just note that $$x^2-4=(x-2)(x+2)=\left(\sqrt{x}-\sqrt2\right)\left(\sqrt{x}+\sqrt2\right)(x+2)‌​\;.$$ What tools do you have available for the first one if you don’t have l’Hospital’s rule? –  Brian M. Scott Jan 12 '13 at 18:59
@Steve Just rewrite $\ln(x/2)$ as $\ln(x) - \ln(2)$ and use the definition of derivative. –  Erick Wong Jan 12 '13 at 19:18