Question

find the value of $\underset{x\to \infty }{lim}\frac{2x^2-3x+11}{7x^3+8x^2+15x+6}$

• A. 0
• B.
• C.
• D. 11

Answer 1

The correct option is A

0

We have, $\underset{x\to \infty }{lim}\frac{2x^2-3x+11}{7x^3+8x^2+15x+6}$

Degree of polynomial in numerator =2

Degree of polynomial in denminator=3

Here, we divide the numberator and denominator by higher degree of polynomial means by $x^3$

so,we have $\underset{x\to \infty }{lim}\frac{\frac{2x^2-3x+11}{x^3}}{\frac{7x^3+8x^2+15x+6}{x^3}}$

= $\underset{x\to \infty }{lim}\frac{\frac{2}{x}-\frac{3}{x^2}+\frac{11}{x^3}}{7+\frac{3}{x}+\frac{15}{x^2}\frac{6}{x^3}}$

=$\frac{0-0+0}{7+0+0+0}=\frac{0}{7}=0$