Question

$\underset{n\to \infty }{lim}\frac{(2n+1{)}^4-(n-1{)}^4}{(2n+1{)}^4+(n-1{)}^4}=$

• A. $\frac{15}{17}$
• B. $\frac{17}{15}$
• C. $0$
• D. $\infty$

Answer 1

The correct option is A $\frac{15}{17}$

$\underset{n\to \infty }{lim}\frac{(2n+1{)}^4-(n-1{)}^4}{(2n+1{)}^4+(n-1{)}^4}$
$\underset{n\to \infty }{lim}\frac{n^4\left[\right(2+\frac{1}{n}{)}^4-(1-\frac{1}{n}{)}^4]}{n^4\left[\right(2+\frac{1}{n}{)}^4+(1-\frac{1}{n}{)}^4]}$
$\underset{n\to \infty }{lim}\frac{2^4-1}{2^4+1}=\frac{15}{17}$