Question

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

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

Answer 1

Answer: (A)

$5$

$\underset{n\to \infty }{lim}\frac{{23}^{n+1}-{35}^{n+1}}{{2.3}^n+{3.5}^n}$

$=\underset{n\to \infty }{lim}\frac{{63}^n-{15.5}^n}{{2.3}^n+{3.5}^n}$

divide No 2 do by $5^n$

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