How can I solve this equation of functions? I need to find two functions continuous on $(-a, a)$, such that:


*

*$f(0) = 0$

*$f'(0) = 0$

*$g(0) = 0$

*$\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} \not= 0$

*The functions are not both writable as $x^n$, where n is a real number (one can be, but not both).

*The $\sin$ and $\cos$ functions are not used


How can I do this?
 A: A solution that uses no monomials is
$$f(x)=e^x-x-1\qquad g(x)=e^x-1$$
The limit in question is $\frac{1}{2}$.
If you're really a stickler, you might want to try to avoid $e^x$ [given its expressability in terms of $\sin(x)$ and $\cos(x)$], and here you can modify mixedmath's answer with
$$f(x)=x^3+x^2 \qquad g(x)=x^2+x$$
Here the limit is 1.
A: If you play around with exponents of monomials, then you're looking for $f(x) = x^n$ and $g(x) = x^m$. By your conditions, you require $n \geq 2, m \geq 1$.
Looking at the ratio 
$$ \frac{f(x)g'(x)}{(f(x)^2} = \frac{mx^{n+m-1}}{x^{2m}} = m x^{n-m-1},$$
you want a solution to $n-m-1 = 0$, or $n = m+1$.
For instance, $f(x) = x^3$ and $g(x) = x^2$ works. So does $f(x) = x^4$ and $g(x) = x^3$, and more generally $f(x) = x^m, g(x) = x^{m+1}$ for any $m \geq 2$.

(When I first answered, I misread the constraint that they both cannot be monomials)
To account for them not both being monomials, it is perhaps easiest to base a response off the work above and incorporate knowledge from series expansion. For instance, for $x$ small, $e^x \approx 1 + x + x^2/2 + O(x^3)$. Correspondingly, $e^x - 1 \approx x + x^2/2 + O(x^3)$.
Then we can adjust $f(x)$ in the start of this answer to a similar function $\widetilde{f}(x)$ (or $g$ into $\widetilde{g}(x)$, but I happen to choose the former) by choosing $\widetilde{f}(x) = e^{mx} - 1$ in place of $f(x) = x^m$.
(This works because for small $x$, $e^{mx} - 1 \approx x^m$).
So an infinite family of choices are $\widetilde{f}(x) = e^{mx}-1$ and $g(x) = x^{m+1}$.
A: $$f(x)=x^2\;,\;\;g(x)=xe^x{}{}{}{}{}{}{}{}{}{}{}$$
A: Let
$$ f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \sum_{n=0}^{\infty} b_{n} \, x^{n} $$
then by the given conditions, $f(0) = 0$, $f'(0) = 0$, $g(0) = 0$, it is found that $b_{0} = 0$, $a_{0} = a_{1} = 0$ and lead to
$$ f(x) = \sum_{n=1}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \sum_{n=2}^{\infty} b_{n} \, x^{n}. $$
There is a lot of freedom to choose the coefficients $a_{n}$ and $b_{n}$.
What remains is: 


*

*$\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} \not= 0$

*The functions are not both writable as $x^n$, where n is a real number (one can be, but not both).


In this view let $b_{n} = \frac{(-1)^{n+1}}{n}$ such that $g(x) = \ln(1+x)$. Now,
\begin{align}
\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} &= \lim_{x \to 0} \frac{(a_{2} \, x^{2} + \cdots + a_{n} \, x^{n} + \cdots)}{(1+x) \, \ln^{2}(1+x) }  \, \to \frac{0}{0} \\
&= \lim_{x \to 0} \frac{(2 \, a_{2} \, x + \cdots + n \, a_{n} \, x^{n-1} + \cdots)}{2  \, \frac{\ln(1+x)}{1+x} }  \, \to \frac{0}{0} \\
&= \lim_{x \to 0} \frac{(2 \, a_{2} + \cdots + n(n-1) \, a_{n} \, x^{n-2} + \cdots)}{2 \, \frac{1}{1+x} }  \, \to a_{2} \neq 0 \\
\end{align}
This allows for 
$$ f(x) = \sum_{n=2}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \ln(1+x). $$
