How to determine the critical exponent of the function $f(x)=Ax^{1/2}+Bx^{1/4}+Cx$?

Question

In the book Statistical Mechanics of Phase Transitions written by J.M. Yeomans there is a set of exercises*, where the objective is to find the critical exponent of specific functions.

The critical exponent ($\lambda$) of a real valued function ($f$) is defined as the following: \begin{equation} \lambda := \lim_{x \to 0} \frac{ \ln(|f(x)|) } { \ln(|x|) } \end{equation}

In my case, I have some trouble to determine the value of the limit for the following function ($f$):

\begin{equation} f(x)=Ax^{1/2}+Bx^{1/4}+Cx \mid A,B,C \in \mathbb{R} \end{equation}

The first method that I tried was to solve the limit in a "classical way". \begin{equation} \begin{split} \lambda &= \lim_{x \to 0} \frac{ \ln(|f(x)|) } { \ln(|x|) } = \lim_{x \to 0} \frac{ \ln(|Ax^{1/2}+Bx^{1/4}+Cx|) } { \ln(|x|) } \\ & = \lim_{x \to 0} \frac{ \ln(|x(Ax^{-1/2}+Bx^{-3/4}+C)|) } { \ln(|x|) } = \lim_{x \to 0} \frac{ \ln(|x| \cdot |Ax^{-1/2}+Bx^{-3/4}+C|) } { \ln(|x|) } \\ & = \lim_{x \to 0} \frac{ \ln(|x|) + \ln(|Ax^{-1/2}+Bx^{-3/4}+C|) } { \ln(|x|) } = 1 + \lim_{x \to 0} \frac{ \ln(|Ax^{-1/2}+Bx^{-3/4}+C|) } { \ln(|x|) } \end{split} \end{equation}

Unfortunately, I could not get anything from it...

After this fail, I tried to rewrite the limit in another form. \begin{equation} \begin{aligned} \lambda = \lim_{x \to 0} \frac{ \ln(|f(x)|) } { \ln(|x|) } &\Leftrightarrow \lambda \lim_{x \to 0} \ln(|x|) = \lim_{x \to 0} \ln(|x|^{\lambda}) = \lim_{x \to 0} \ln(|x^{\lambda}|) = \lim_{x \to 0} \ln(|f(x)|)\\ &\Leftrightarrow \lim_{x \to 0} \exp({\ln(|x^{\lambda}|)}) = \lim_{x \to 0} |x^{\lambda}| = \lim_{x \to 0} |f(x)|\\ &\Leftrightarrow 1 = \lim_{x \to 0} \frac{|f(x)|}{|x^{\lambda}|} \end{aligned} \end{equation} By plugging in the function, I get: \begin{equation} \begin{aligned} 1 &= \lim_{x \to 0} \frac{|Ax^{1/2}+Bx^{1/4}+Cx|}{|x^{\lambda}|} = \lim_{x \to 0} \frac{|x^{\lambda} \cdot(Ax^{1/2-\lambda}+Bx^{1/4-\lambda}+Cx^{-\lambda})|}{|x^{\lambda}|}\\ &= \lim_{x \to 0} \frac{|x^{\lambda} \cdot(Ax^{1/2-\lambda}+Bx^{1/4-\lambda}+Cx^{-\lambda})|}{|x^{\lambda}|} = \lim_{x \to 0} \frac{|x^{\lambda}|}{|x^{\lambda}|} \cdot {|Ax^{1/2-\lambda}+Bx^{1/4-\lambda}+Cx^{-\lambda}|}\\ &= \lim_{x \to 0} {|Ax^{1/2-\lambda}+Bx^{1/4-\lambda}+Cx^{-\lambda}|} \end{aligned} \end{equation} From this point, I still do not know where to go, I have the feeling that I should have more informations about the constants $A$, $B$, and $C$, if I really want to solve the limit.

I would be interested to know if there is a nice way to find the critical exponent of this function.

Moreover, I am open to any suggestion of book treating in a rigourous way this kind of problem, since I do not have enough experience to solve them.

Thank you in advance for your help.

P.S.: In the book, it is written that the critical exponent of $f(x)=Ax^{1/2}+Bx^{1/4}+Cx$ is 1/4.

*YEOMANS, Julia Mary. Oxford Science Publications: Statistical Mechanics of Phase Transitions. 1st Edition. Clarendon Press. Page 31.

Answer 1

First, from the definition of $f$, you have $x\ge 0$. For simplicity, let's assume that in a small region around $0$ you have positive $f(x)$. For the negative case, you will just need to change a sign. Then $|x|=x$ and $|f(x)|=x$. Now use l'Hospital:

$$\lim_{x\to 0}\frac{ln(f(x))}{ln(x)}=\lim_{x\to 0}\frac{\frac{f'(x)}{f(x)}}{\frac 1x}=\lim_{x\to 0}\frac{xf'(x)}{f(x)}=\lim_{x\to 0}\frac{x(\frac 12Ax^{-1/2}+\frac14Bx^{-3/4}+C)}{Ax^{1/2}+Bx^{1/4}+Cx}=\lim_{x\to 0}\frac{\frac 12Ax^{1/2}+\frac14Bx^{1/4}+Cx}{Ax^{1/2}+Bx^{1/4}+Cx}=\lim_{x\to 0}\frac{\frac14Bx^{1/4}(2\frac AB x^{1/4}+1+4\frac CB x^{3/4})}{Bx^{1/4}(\frac AB x^{1/4}+1+\frac CB x^{3/4})}=\frac 14$$

Answer 2

Rewrite

$$ \frac{\ln |f(x)|}{\ln |x|} $$

as

$$ \frac{\ln |Bx^{1/4}|}{\ln |x|} + \frac{\ln\left|1 + \frac{A}{B}x^{1/4}+\frac{C}{B}x^{3/4}\right|}{\ln |x|} $$

As $x \to 0$, the argument of natural log in the second numerator goes to $1$.