Question

find the value of $\underset{x\to \infty }{lim}\frac{2x^3-3x^2+5x+6}{7x^2+8x+15}$

• A. 0
• B.
• C.
• D. 6

Answer 1

The correct option is B

Here,we have $\underset{x\to \infty }{lim}\frac{2x^3-3x^2+5x+6}{7x^2+8x+15}$

Degree of polynomial in numerator and denominator is NOT same.

Degree of polynomial in numerator =3

Degree of polynomial in denminator=2

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

so,we have

$\underset{x\to \infty }{lim}\frac{\frac{2x^3-3x^2+5x+6}{x^3}}{\frac{7x^2+8x+15}{x^3}}$

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

=$\frac{2-0+0+0}{0}=\frac{2}{0}=\infty$