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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.

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    @AndréNicolas The function isnt even well-defined ^.^ for x=0, we have $0/0$ which is indeterminate, however, the interval is set from $[0,\infty)$. If it were defined from $\mathbb{R}\rightarrow\mathbb{R}$, f(x)= \begin{cases} \frac{x}{1-e^x} &\mbox{if } x\neq 0 \\ 1 & \mbox{else}\end{cases} , (a) and (c) are true. EDIT: nvm, cant read >_< – 2012-12-10

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Hints:

a) Use L'Hospital's Rule, or some other method, to show that the limit from the right is $1$.

b) The function is $\gt x$.

c) $e^x\ge 1+x$.

d) Needs no hint.