Question

$\underset{x\to \infty }{lim}\frac{5x^2-7x+6}{x^2+8}=$

• A. $\frac{6}{8}$
• B. $\frac{8}{6}$
• C. $5$
• D. $\frac{7}{8}$

Answer 1

The correct option is C $5$

The value of $\underset{x⟶\infty }{lim}\frac{5x^2-7x+6}{x^2+8}$ is

$=\underset{x⟶\infty }{lim}\frac{x^2(5-\frac{7}{x}+\frac{6}{x^2})}{x^2(1+\frac{8}{x^2})}$

$=\underset{x⟶\infty }{lim}\frac{5-\frac{7}{x}+\frac{6}{x^2}}{1+\frac{8}{x^2}}$

$=\frac{5-0+0}{1+0}$

$=5$