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 equal to

• A. $5$
• B. $3$
• C. $1$
• D. zero

Answer 1

The correct option is D zero

Now,

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

$=\underset{n\to \infty }{lim}\frac{{5.5}^n+3^n-4^n}{5^n+2^n+27.9^n}$

$=\underset{n\to \infty }{lim}\frac{5.{\left(\frac{5}{9}\right)}^n+{\left(\frac{3}{9}\right)}^n-{\left(\frac{4}{9}\right)}^n}{{\left(\frac{5}{9}\right)}^n+{\left(\frac{2}{9}\right)}^n+27}$ [ Dividing the numerator and the denominator by $9^n$ we get]

$=\frac{5.0+0-0}{0+0+27}$ [ Since $\underset{n\to \infty }{lim}{a}^n=0$ for $a<1$]

$=0$.