Question

${lim}_{n\to \infty }\frac{4n^3-3n^2+5}{6n^3+4n^2-3}$ is equal to

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $1$
• D. $0$

Answer 1

The correct option is B $\frac{2}{3}$

$\underset{n\to \infty }{lim}\frac{4x^3-3x^2+5}{6x^3+4x^2-3}$

$\underset{n\to \infty }{lim}\frac{n^3(4-\frac{3}{n}+\frac{5}{n^3})}{n^3(6+\frac{4}{n}-\frac{3}{n^3})}$

$\underset{\frac{1}{n}\to 0}{lim}\frac{4-\frac{3}{n}+\frac{5}{n^3}}{6+\frac{4}{n}-\frac{3}{n^3}}=\frac{4}{6}=\frac{2}{3}$.