Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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

If we substitute $x=\infty$ in the above expression it becomes $\frac{\infty }{\infty }$ form.

Degree of polynomial in numerator and denominator is same which is 2. We can divide numerator and denominator by $x^2$.
So, when we substitute $x=\infty$ in $\frac{1}{x}$&$\frac{1}{x^2}$, it becomes zero.

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

$=\frac{2-0+0}{7+0+0}$

$=\frac{2}{7}$