[tex]\displaystyle\\\lim_{n\to\infty}\left(\dfrac{n+4}{n+3}\right)^{5-2n}=\lim_{n\to\infty}\left(\dfrac{n+3+1}{n+3}\right)^{5-2n}=\lim_{n\to\infty}\left(1+\dfrac{1}{n+3}\right)^{5-2n}=\\\\=\lim_{n\to\infty}\left(\left(1+\dfrac{1}{n+3}\right)^{n+3}\right)^{\tfrac{5-2n}{n+3}}=e^{\displaystyle \lim_{n\to\infty}\frac{5-2n}{n+3}}}}=e^{\displaystyle \lim_{n\to\infty}\frac{\frac{5}{n}-2}{1+\frac{3}{n}}}}}=\\\\=e^{\dfrac{0-2}{1+0}}=e^{-2}=\dfrac{1}{e^2}[/tex]
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[tex]\displaystyle\\\lim_{n\to\infty}\left(\dfrac{n+4}{n+3}\right)^{5-2n}=\lim_{n\to\infty}\left(\dfrac{n+3+1}{n+3}\right)^{5-2n}=\lim_{n\to\infty}\left(1+\dfrac{1}{n+3}\right)^{5-2n}=\\\\=\lim_{n\to\infty}\left(\left(1+\dfrac{1}{n+3}\right)^{n+3}\right)^{\tfrac{5-2n}{n+3}}=e^{\displaystyle \lim_{n\to\infty}\frac{5-2n}{n+3}}}}=e^{\displaystyle \lim_{n\to\infty}\frac{\frac{5}{n}-2}{1+\frac{3}{n}}}}}=\\\\=e^{\dfrac{0-2}{1+0}}=e^{-2}=\dfrac{1}{e^2}[/tex]
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