How to use $\epsilon$-$\delta$ to show that following is less than $\epsilon$?

Question

I try to $\epsilon$-$\delta$ to show that if I fix $\alpha$ small enough, for $\forall \epsilon>0$, there exists a $N$ such that $k>N$, I want $$\frac{c}{\alpha}\exp(-2^k)+\exp\left(-\frac{1}{\alpha}\right)\le \epsilon$$ where $c>0$ is a constant.

I think I just need to let $\frac{c}{\alpha}\exp(-2^k)<\epsilon$ which is $$2^k>\log\left(\frac{c}{\alpha \epsilon}\right)$$ But I do not know how to deal with $\exp\left(-\frac{1}{\alpha}\right)$. This term is small enough...

The statement:

For $\forall \epsilon>0$, we set $\epsilon-\exp\left(-\frac{1}{\alpha}\right)>0$, then we can find $$2^k>log(\frac{c}{\alpha(\epsilon-\exp(-1/\alpha))})$$ such that $$\frac{c}{\alpha}\exp(-2^k)+\exp\left(-\frac{1}{\alpha}\right)\le \epsilon$$

Answer 1

As you have stated the problem, the inequality doesn't hold when

$$\varepsilon -\exp\left(-\frac{1}{\alpha}\right)\le 0$$

since, assuming $\frac c \alpha>0$, $\forall k, \: \frac{c}{\alpha}\exp(-2^k)>0$.

The problem should be stated as follows

$\forall \epsilon>0$, there exist $\alpha>0$ small enough and $N$ such that $\forall k,\:k>N$ $$\frac{c}{\alpha}\exp(-2^k)+\exp\left(-\frac{1}{\alpha}\right)\le \epsilon$$ where $c>0$ is a constant.

Then fixed $\varepsilon>0$, we need to set $\alpha$ such that

$$\varepsilon -\exp\left(-\frac{1}{\alpha}\right)=\varepsilon_2 >0$$

and, assuming $\frac c \alpha>0$, we obtain

$$\frac{c}{\alpha}\exp(-2^k)+\exp\left(-\frac{1}{\alpha}\right)\le \varepsilon \iff \frac{c}{\alpha}\exp(-2^k)\le \varepsilon_2 \iff k\ge \frac1{\log 2}\log\left(\log\left(\frac{c}{\alpha \varepsilon_2}\right)\right)$$

The case with $\alpha <0$, if necessary, can be discussed in a similar way.