By using the definition of $e$, prove that $\lim_{x\to\infty} \left(\frac{3+2x}{5+2x}\right)^x = \frac1e$

Question

By using the definition of $e$, prove that: $$\lim_{x\to\infty} \left(\frac{3+2x}{5+2x}\right)^x = \frac1e.$$

I can solve it by using l'hospital rule. But how to use definition of $e$ to show it?

Answer 1

HINT:

$$\left(\dfrac{3+2x}{5+2x}\right)^x=\dfrac{\left(1+\dfrac3{2x}\right)^x}{\left(1+\dfrac5{2x}\right)^x}=?$$

Answer 2

The definition of $e$ is \begin{eqnarray} e^x = \lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n \end{eqnarray} Try to get it to the format. I will edit it for the full solution if there are more problems arising.

Answer 3

$$\lim_{x\rightarrow\infty}\left(\frac{3+2x}{5+2x}\right)^x=\lim_{x\rightarrow\infty}\left(1+\frac{3+2x}{5+2x}-1\right)^x=\lim_{x\rightarrow\infty}\left(1-\frac{2}{5+2x}\right)^x=$$ $$=\lim_{x\rightarrow\infty}\left(1-\frac{2}{5+2x}\right)^{-\frac{5+2x}{2}\cdot\frac{-2x}{5+2x}}=\left(\lim_{x\rightarrow\infty}\left(1-\frac{2}{5+2x}\right)^{-\frac{5+2x}{2}}\right)^{\lim\limits_{x\rightarrow\infty}\frac{-2x}{5+2x}}=\frac{1}{e}.$$