Evaluation of the limit, $\lim \limits_{x\rightarrow\infty} \left(\frac{20x}{20x+4}\right)^{8x}$, 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?
 A: HINT:
Always start by simplifying
$$\left(\frac{20x}{20x+4}\right)^{8x}=\left(\frac{20x+4}{20x}\right)^{-8x}=\left(1+\frac{1}{5x}\right)^{-8x}$$
perhaps a substitution helps... 
A: HINT: 
\begin{align*}
\lim_{x \to \infty} \biggl(\frac{20x}{20x+4}\biggr)^{8x} &=\lim_{x \to \infty} \biggl( 1 - \frac{4}{20x+4}\biggr)^{8x} \\ &=\lim_{x \to \infty} \biggl(1 - \frac{4}{20x+4}\biggr)^{\frac{20x+4}{4} \cdot \frac{8x}{20x+4} \cdot 4}&\end{align*}
Now use the fact that $$\lim_{n \to \infty} \biggl(1 + \frac{1}{n}\biggr)^{n} =e$$
ADDED: After simplyfing you should get this. $$e^{\displaystyle \lim_{x \to \infty} - \frac{8x}{20x+4} \cdot 4} = e^{-8/5}$$  
A: Hint: Consider taking logarithms before applying L'Hospital's rule.
A: 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
$$1/\alpha^5 = \lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{40x}=
\left(\lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{5x}\right)^8 = e^8.$$
A: We can calculate Ln() of the function and result is exp() of the answer.
$$ \lim_{x \rightarrow \infty} \left[ ( \frac{20x}{20x+4})^{8x} \right] =  \lim_{x \rightarrow \infty} \left[ ( \frac{5x}{5x+1})^{8x} \right]$$
$$  Ans= \lim_{x \rightarrow \infty} \left[ 8x. Ln ( \frac{5x}{5x+1}) \right] = \infty . 0 $$
$$ Ans=8 \lim_{x \rightarrow \infty} \left[ \frac{ Ln ( \frac{5x}{5x+1})}{\frac{1}{x}} \right] = \frac{0}{0} \space  \rightarrow  \space Ambiguity in Mathematics  $$
we can use Hopital:
$$ Hopital : \frac{derivative \space of  \space the  \space  Numerator}{derivative  \space  of  \space  the \space  denominator}$$
$$  Ans=8\lim_{n \rightarrow \infty}\left[  \frac{ \frac{1}{x(5x+1)}}{\frac{-1}{x^2}} \right]=8\lim_{n \rightarrow \infty} \left[ \frac{-x}{5x+1} \right]=8(\frac{-1}{5}) $$
therefor : 
$$  \lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x} = e^{Ans}=e^{\frac{-8}{5}} $$
