pprove that for alll $\displaystyle 0<c<\frac{{1}}{2}$ there is x>0 so $\displaystyle \frac{1}{x}-\frac{1}{e^{x}-1}=c$

i have tried $\displaystyle f(x)=\frac{1}{x}-\frac{1}{e^{x}-1}$

limit of f(x) x->infinity = zero

limit of f(x) x->0 = pi/2

so in similar problems i have defined

g(x)= 0.5 when x=0

g(x)= f(x) when 0<x<infinity

g(x)=0 when x=infinity

and then i have to point one is 0 the other is 1/2

and i do mean value theorem

but its not legat to definne such function

infinity is not a point,what to do?