Proving that $\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=e^{-2}$ without using de l‘Hôspital

Question

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

But how do I prove this without using de l'Hôspital twice?

Answer 1

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

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

Answer 2

Hint : $\frac{2n-1}{2n+3} = 1 - \frac{4}{2n+3}$. Now change the variable and you've got a necessary.

Answer 3

Without change of variables, note that

$$ \left ( \frac{2n-1}{2n+3} \right ) ^n =\left ( \frac{2n+3-4}{2n+3} \right ) ^n=\left ( 1+\frac{-4}{2n+3} \right ) ^n=\sqrt{\left ( 1+\frac{-4}{2n+3} \right ) ^{2n+3}\left ( 1+\frac{-4}{2n+3} \right ) ^{-3}}\to \sqrt{e^{-4}\cdot1}=\frac1{e^2}$$

Answer 4

It's the case of $\lim\limits_{\infty} a_n^{b_n}$ when $a_n \to 1$ and $b_n \to \infty$, so: $$\lim\limits_{\infty}a_n^{b_n}=\exp(\lim\limits_{\infty}b_n*(a_n-1))$$ So: $$b_n*(a_n-1)=n*\frac{-4}{2n+3}=\frac{-4n}{2n+3}\to\frac{-4}{2}=-2$$ So the original limit is: $$\exp(-2)=e^{-2}=\frac{1}{e^2}$$