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Question Number 139059 by sahnaz last updated on 21/Apr/21

Answered by mr W last updated on 21/Apr/21

$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{2\times {\left(\frac{2}{4}\right)}^n+9\times {\left(\frac{3}{4}\right)}^n+64}{{\left(\frac{2}{4}\right)}^n+{\left(\frac{3}{4}\right)}^n+1}$$$$=\frac{2\times 0+9\times 0+64}{0+0+1}$$$$=\frac{64}{1}$$$$=64$$

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