Question

$\underset{n\to \infty }{lim}\frac{5^{n+1}+3^n-2^{2n}}{5^n+2^n+3^{2n+3}},n\in N$ is :

• A. $5$
• B. $3$
• C. $1$
• D. $Zero$

Answer 1

The correct option is A $5$

$\underset{x\to \infty }{lim}$ $\frac{5^{n+1}+3^n-2^{2n}}{5^n+2^n+3^{2n}.3^3}$

$=\underset{x\to \infty }{lim}$ $\frac{5^n(5+{\left(3/5\right)}^n-2^2.{\left(2/5\right)}^n)}{5^n(1+{\left(2/5\right)}^n+3^2.{\left(3/5\right)}^n).3^3}$

$=\frac{5}{1}=5$