<s>[INST] 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?
 [/INST] Without change of variables, note that
$$ \left ( \frac{2n-1}{2n+3} \right ) ^n =\left ( \frac{2n+3-4}{2n+3} \right ) ^n=\left ( 1+\frac{-4}{2n+3} \right ) ^n=\sqrt{\left ( 1+\frac{-4}{2n+3} \right ) ^{2n+3}\left ( 1+\frac{-4}{2n+3} \right ) ^{-3}}\to \sqrt{e^{-4}\cdot1}=\frac1{e^2}$$
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