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Question Number 57866 by Mikael_Marshall last updated on 13/Apr/19

$$\underset{n\to \mathrm{\infty }}{lim}\frac{3^{n+2}-2.5^{n+1}}{3^n-2.5^{n-1}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{3^n\times 3^2-2\times 5^n\times 5}{3^n-2\times 5^n\times 5^{-1}}$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{9\times 3^n-10\times 5^n}{3^n-0.4\times 5^n}$$$$5^n>3^n$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{9\times {\left(\frac{3}{5}\right)}^n-10}{{\left(\frac{3}{5}\right)}^n-0.4}$$$$=\frac{9\times 0-10}{0-0.4}$$$$=\frac{10}{0.4}=\frac{100}{4}=25$$

Commented by Mikael_Marshall last updated on 13/Apr/19

$$yesSir$$

Answered by Kunal12588 last updated on 13/Apr/19

$$let{3}^n-2\cdot 5^{n-1}=k$$$$\Rightarrow 3^n=k+2\cdot 5^{n-1}and2\cdot 5^{n-1}=3^n-k$$$$Now,$$$$3^{n+2}-2\cdot 5^{n+1}$$$$=3^n\cdot 3^2-2\cdot 5^{n-1}\cdot 5^2$$$$=3^{2}k+2\cdot 3^2\cdot 5^{n-1}-2\cdot 5^{n-1}\cdot 5^2$$$$=3^{2}k+2\cdot 5^{n-1}(3^2-5^2)$$$$=3^{2}k-16(3^n-k)$$$$=25k-16\cdot 3^n$$$$\therefore \underset{n\to \mathrm{\infty }}{lim}\frac{3^{n+2}-2\cdot 5^{n+1}}{3^n-2\cdot 5^{n-1}}$$$$=\underset{n\to \mathrm{\infty }}{lim}\frac{25k-16\cdot 3^n}{k}$$$$=\underset{n\to \mathrm{\infty }}{lim}(25-\frac{16\cdot 3^n}{3^n-2\cdot 5^{n-1}})$$$$dividenumeratoranddenominatorwith{5}^n$$$$assirTanmayshowed$$$$=25-0$$$$=25$$

Commented by Mikael_Marshall last updated on 13/Apr/19

$$thanksSir$$

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