Question

$\underset{n\to \infty }{lim}\frac{{2.3}^{n+1}-{3.5}^{n+1}}{{2.3}^n+{3.5}^n}=$

• A. $5$
• B. $\frac{1}{5}$
• C. $-5$
• D. $0$

Answer 1

The correct option is C $-5$

$\underset{n\to \infty }{lim}\frac{{2.3}^{n+1}-{3.5}^{n+1}}{{2.3}^n+{3.5}^n}=\underset{n\to \infty }{lim}\frac{{6.3}^n-{15.5}^n}{{2.3}^n+{3.5}^n}$

$=\underset{n\to \infty }{lim}\frac{6.{\left(\frac{3}{5}\right)}^n-15}{2.{\left(\frac{3}{5}\right)}^n+3}$, dividing Nr and Dr by $5^n$

$=\frac{-15}{3}=-5$, since $\frac{3}{5}<1$