Question

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

• A. $2$
• B. $\infty$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

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

$=\underset{n\to \infty }{lim}\frac{n\times n(2-1/n)(3+5/n)}{n\times n(1-1/n)(3+1/n)(3^n+2^n)}[n\to \infty ,\frac{1}{n}\to 0]$

$=\frac{(2-0)(3+0)}{(1-0)(3+0)(3^n+2^n)}=\frac{6}{3}\times \frac{1}{\infty }=0$