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

1. nikvist

$\lim_{n\rightarrow\infty}\frac{3^{n+1}+2^{n+2}}{3^{n+1}+2^{n-2}}=\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$

2. anonymous

I 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

3. anonymous

Its 1 because limit n tends to infinity x^n is 0 if 0<x<1.