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anonymous
 4 years ago
What will be the value of lim (3^(n+1) + 2^(n+2))/ (3^(n+1)+2^(n2)) wher n tends to
infinity??
anonymous
 4 years ago
What will be the value of lim (3^(n+1) + 2^(n+2))/ (3^(n+1)+2^(n2)) wher n tends to infinity??

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n\rightarrow\infty}\frac{3^{n+1}+2^{n+2}}{3^{n+1}+2^{n2}}=\lim_{n\rightarrow\infty}\frac{3\cdot3^{n}+4\cdot 2^{n}}{3\cdot3^{n}+1/4\cdot 2^{n}}=\]\[=\lim_{n\rightarrow\infty}\frac{3+4\cdot (2/3)^{n}}{3+1/4\cdot (2/3)^{n}}=1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did not get how you got answer as 1. And also the options for this question are: A) 9 B) 3/2 C) 2/3 D) infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its 1 because limit n tends to infinity x^n is 0 if 0<x<1.
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