[tex]\displaystyle\\\lim_{n\to\infty}\dfrac{5^{n-1}+(-3)^n}{5^{n+1}+(-3)^{n+3}}=\lim_{n\to\infty}\dfrac{5^n\cdot 5^{-1}+(-3)^n}{5^n\cdot 5+(-3)^n\cdot (-3)^3}=\\\\=\lim_{n\to\infty}\dfrac{5^n\cdot (-3)^n\left(\dfrac{1}{5\cdot(-3)^n}+\dfrac{1}{5^n}\right)}{5^n\cdot(-3)^n\left(\dfrac{5}{(-3)^n}-\dfrac{27}{5^n}\right)}=\lim_{n\to\infty}\dfrac{\dfrac{1}{5\cdot(-3)^n}+\dfrac{1}{5^n}}{\dfrac{5}{(-3)^n}-\dfrac{27}{5^n}}=[/tex]
[tex]\displaystyle=\lim_{n\to\infty}\dfrac{\dfrac{1}{5\cdot(-3)^n}}{\dfrac{5}{(-3)^n}}=\lim_{n\to\infty}\dfrac{1}{5\cdot(-3)^n}\cdot\dfrac{(-3)^n}{5}=\dfrac{1}{25}[/tex]
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[tex]\displaystyle\\\lim_{n\to\infty}\dfrac{5^{n-1}+(-3)^n}{5^{n+1}+(-3)^{n+3}}=\lim_{n\to\infty}\dfrac{5^n\cdot 5^{-1}+(-3)^n}{5^n\cdot 5+(-3)^n\cdot (-3)^3}=\\\\=\lim_{n\to\infty}\dfrac{5^n\cdot (-3)^n\left(\dfrac{1}{5\cdot(-3)^n}+\dfrac{1}{5^n}\right)}{5^n\cdot(-3)^n\left(\dfrac{5}{(-3)^n}-\dfrac{27}{5^n}\right)}=\lim_{n\to\infty}\dfrac{\dfrac{1}{5\cdot(-3)^n}+\dfrac{1}{5^n}}{\dfrac{5}{(-3)^n}-\dfrac{27}{5^n}}=[/tex]
[tex]\displaystyle=\lim_{n\to\infty}\dfrac{\dfrac{1}{5\cdot(-3)^n}}{\dfrac{5}{(-3)^n}}=\lim_{n\to\infty}\dfrac{1}{5\cdot(-3)^n}\cdot\dfrac{(-3)^n}{5}=\dfrac{1}{25}[/tex]