Differentiation Question 107
Question: $ \underset{n\to \infty }{\mathop{\lim }}\frac{{5^{n+1}}+3^{n}-2^{2n}}{5^{n}+2^{n}+{3^{2n+3}}} $ is equal to
Options:
A) 5
B) 3
C) 1
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \underset{n\to \infty }{\mathop{\lim }}\frac{{5^{n+1}}+3^{n}-2^{2n}}{5^{n}+2^{n}+{3^{2n+3}}}=\underset{n\to \infty }{\mathop{\lim }}\frac{{{5.5}^{n}}+3^{n}-4^{n}}{5^{n}+2^{n}+{{27.9}^{n}}} $
$ =\underset{n\to \infty }{\mathop{\lim }}\frac{5.\frac{5^{n}}{9^{n}}+\frac{3^{n}}{9^{n}}-\frac{4^{n}}{9^{n}}}{\frac{5^{n}}{9^{n}}+\frac{2^{n}}{9^{n}}+27}=\frac{0+0-0}{0+0+27}=0. $
BETA
