I was thinking about the problem that says:
Let $f:\mathbb [0,\infty)\rightarrow \mathbb R$ be defined by $f(x)=\frac{x}{1-e^{-x}}$ if $x>0$ and $f(0)=0$.
Then the function is
(a)continuous at $x=0,$
(b)bounded,
(c)increasing,
(d)zero for at least one $x>0.$
Here is my attempts:
For $x>0,f(x)=\frac{x}{1-e^{-x}}=\frac{1}{1-x/2!+x^{2}/3!-...}=\frac{1}{1-p},where p=x/2!-x^{2}/3!+...$ and hence $f(x)=(1-p)^{-1}=1+p+p^{2}+p^{3}+p^{4}+p^{5}+.....$ which shows that f is not bounded. From the series representations, it appears that f is increasing. I can also show that f is not continuous at $x=0.$ So, I can eliminate the choices $(a), (b)$. So the answer should be $(c)$.Since $f(x) \neq 0$ for any $x>0$.,option $(d)$ also is not possible. Am I going in the right direction? Please help. Thanks in advance for your time.