I need to solve the following exercise:

**Exercise:** What values of $ alpha $, if applicable, performance $ f = mathcal {O} ( epsilon ^ alpha) $ as $ epsilon downarrow 0 $ Yes

$$

f = dfrac {1} {1 – e ^ epsilon}

$$

**What I have tried:**

I know I can use the following theorem:

*Yes*

$$

lim_ {e downarrow e_o} dfrac {f ( epsilon)} { phi ( epsilon)} = L,

$$

where $ – infty <L < infty $, so $ f = mathcal {O} ( phi) $ as $ epsilon downarrow epsilon_0 $.

So I think I need to evaluate the limit

$$

lim _ { epsilon downarrow 0} dfrac {1} { epsilon ^ alpha (1 – e ^ epsilon)}

$$

and find out for what values of $ alpha $ exists. The limit exists if $ epsilon ^ alpha (1 – e ^ epsilon) $ converges to infinity or any other number that is not $ 0 $. I know that $ 1 – e ^ epsilon a 1 – 1 = 0 $ for $ epsilon downarrow 0 $. Then the limit exists if $ epsilon ^ alpha to infty $ faster than $ 1 – e ^ epsilon a 0 $because in that case $ epsilon ^ alpha (1 – e ^ epsilon) a infty $ Y

$$ lim _ { epsilon to 0} dfrac {1} { epsilon ^ alpha (1-e ^ epsilon)} = 0 $$

Yes $ alpha> 0 $ so $ lim _ { epsilon downarrow 0} epsilon ^ alpha = 0 $ so we know that at least $ alpha leq 0 $ necessary. Yes $ alpha = 0 $ so $ lim _ { epsilon downarrow 0} epsilon ^ alpha = 1 $. Yes $ alpha <0 $ so $ lim _ { epsilon downarrow 0} epsilon ^ alpha = infty $. That's why I think that for all values of $ alpha <0 $ we have that $ f = mathcal {O} ( epsilon ^ alpha) $.

**Question:** If my reasoning about when the limit exists is correct, I think the weak point of my solution so far lies in the fact that I do not really show / know whether it is true or not. $ epsilon ^ alpha to infty $ faster than $ 1 – e ^ epsilon a 0 $. Is this true? And if so, how can I prove what it is?