Is this proof sufficient? I have two functions $f(x)$ and $g(x)$, where $0<x<1$. I want to prove that $f(x)>g(x)$ on the mentioned domain (Plots of the functions verify this).
Proof:
I assume by contradiction that $f(x)\leq g(x)$. Then, after simplifying this condition, I arrive at a new condition
$h(x)\leq x\quad (1)$,
Then, calculating the value of $h$ at the endpoints of the domain, I get:
$\lim_{x\to 0} \, h(x)=0 \quad and \quad \lim_{x\to 1} \, h(x)=5$
Now, for the equal sign in Eq. (1), for $x=1$, we have $5=1$ which is a contradiction. And, if the inequality sign holds in (1), it yields the range of $h$ as $(0,1)$, while from the limit above, we know that $h(1)=5$ which is greater than $1$.
Then, can we conclude that the proposition is false and $f$ has to be greater than $g$?
Any comment is appreciated.

From what I read I infer that you want to disprove $h(x)\le x$, knowing that $\lim_{x\to 0}h(x)=0$ and $\lim_{x\to1} h(x)=5$. In fact $f$ and $g$, which are not disclosed, play no role here.
Then $\lim_{x\to 0}h(x)=0$ tells us nothing (because $\lim_{x\to 0}x=0$ as well), but $\lim_{x\to1} h(x)=5>1$ is enough to prove that there are values of $x$ such that $h(x)>x$ (take an $\epsilon$ such that $5-\epsilon>1$).
You could as well exhibit a single value of $x$ such that $h(x)>x$.

Anyway, make sure that your transformation from the $f,g$ representation to $h$ does not alter the comparison.
