Proving that $\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=e^{-2}$ without using de l‘Hôspital > $$\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=\frac{1}{e^2}.$$ But how do I prove this without using de l'Hôspital twice?

Knowing that $$e^a =\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^x$$ we have

$$\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=\lim_{n\to\infty}\left ( 1 - \frac{4}{2n+3} \right )^{2n+3\frac{n}{2n+3}}\overset{x=2n+3}{=} \lim_{x\to\infty}\left ( 1 - \frac{4}{x} \right )^{\frac{x}{2}}=e^{-2}.$$