calculus - Evaluation of the limit, $lim limits_{xrightarrowinfty} left(frac{20x}{20x+4}right)^{8x}$, using only elementary methods

I was assisting a TA for an introductory calculus class with the following limit,

$$\lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x}$$

and I came to simple solution which involved evaluating the "reciprocal" limit

$$\lim_{z \rightarrow 0} \left(\frac{1}{1+\frac{z}{5}}\right)^{8/z}$$

by using the Taylor expansion of $\log(1+z)$ around $z=0$. However, the TA claims that the students have not learned about series expansions so that would not be a valid solution for the course. I tried applying L'Hopital's rule, which I was told the class did cover, but I was unsuccessful. As a note I will mention that

$$\lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x} = e^{-8/5}.$$

Any ideas for a solution to this problem using only knowledge from a first quarter (or semester) calculus course which hasn't covered series expansions?

Answer

If they know the definition of $e$ as
$$\lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n,$$
then set $\alpha=\lim_{x\rightarrow\infty}\left(\frac{20x}{20x+4}\right)^{8x}$
and note that
$$1/\alpha^5 = \lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{40x}=
\left(\lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{5x}\right)^8 = e^8.$$