Assume that $x > c > 0 $. Prove that $\displaystyle \inf\left(c/x \frac{e^{c/x}}{e^{c/x} - 1}\right) = 1$ without using limit.
However you are all allowed to use the theorem about the exponential function: The exponential function $e:\mathbb R\to\mathbb R$ has the following properties:
$1.~e^u < \frac1{1-u} $for all $u < 1$
$2.~e^u > 1 + u$ for all $u\in\mathbb R$
$3.~e^u $ is a growing function for all $u\in\mathbb R$
$4.~e^u > 0 $ for all $u\in\mathbb R$
$5.~e^u \cdot e^{-u} = 1 $ for all $u\in\mathbb R$
By using property 2 I get the following expression: $\displaystyle \inf\left(c/x \frac{e^{c/x}}{e^{c/x} - 1}\right) > c/x \frac{{1+c/x}}{e^{c/x}-1} $
But I don't get any further. If you have any hints please let me know.
Here is an argument which uses no limits and does not use L'Hopital's Rule. Only the stated properties of exponential function are uses. As mentioned in the comments the question is equivalent to showing that the infimum of $\frac {te^{t}} {e^{t}-1} $ ove $(0,1)$ is $1$. Now $\frac {te^{t}} {e^{t}-1} >1$ for all $t \in (0,1)$ by property 1). So the infimum is $\geq 1$. Suppose, if possible, the infimum is $1+\epsilon$ for some $\epsilon >0$. Then $\frac {te^{t}} {e^{t}-1} >1+\epsilon$ for all $t \in (0,1)$. This gives $e^{t} (1+\epsilon-t) < 1+\epsilon$. Now use property 2) to get $(1+t) (1+\epsilon-t) < 1+\epsilon$. This simplifies to $\epsilon t <t^{2}$ or $\epsilon <t$ for all $t \in (0,1)$ which is a contradiction.