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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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For the first, the calculation is not quite right. I would rather say use L'Hospital's Rule to show that the limit as $x$ approaches $0$ from the right is $1$. But series correctly handled is good too. –  André Nicolas Dec 10 '12 at 19:16
    
The function seems to be continous in 0. It has a removable singulartiy. –  CBenni Dec 10 '12 at 19:18
    
@CBenni: It would be continuous if $f(0)$ had been defined o be $1$. –  André Nicolas Dec 10 '12 at 19:22
    
@learner: OK, now unbounded. Our function is $\gt x$. –  André Nicolas Dec 10 '12 at 19:27
    
The fourth option? For the function to be $0$, the top would need to be $0$. –  André Nicolas Dec 10 '12 at 19:36

1 Answer 1

up vote 1 down vote accepted

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.

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