Question

Solve :
$\underset{n\to \infty }{Limit}\frac{5^{n+1}+3^n-2^{2n}}{5^n+2^n+3^{2n+3}}=$

Answer 1

$\begin{array}{c}{lim}_{n\to \infty }\frac{5^{n+1}+3^n-2^{2n}}{5^n+2^n+3^{2n+3}}={lim}_{n\to \infty }\frac{5^{n+1}+3^n-4^n}{5^n+2^n+27\cdot 9^n}\\ ={lim}_{n\to \infty }\frac{5{\left(\frac{5}{9}\right)}^n+{\left(\frac{1}{3}\right)}^n-{\left(\frac{4}{9}\right)}^n}{{\left(\frac{5}{9}\right)}^n+{\left(\frac{2}{9}\right)}^n+27}\\ =\frac{0}{27}=0\end{array}$

Hence,

Required answer is $0$.