Question

Evaluate the limit:
$\underset{x\to \infty }{lim}\frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7}$

Answer 1

We have,
$\underset{x\to \infty }{lim}\frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7}$

Dividing numerator and denominator by $x^3$, we get

$=\underset{x\to \infty }{lim}\frac{\left(\frac{3x^3-4x^2+6x-1}{x^3}\right)}{\left(\frac{2x^3+x^2-5x+7}{x^3}\right)}$

$=\underset{x\to \infty }{lim}\frac{(3-\frac{4}{x}+\frac{6}{x^2}-\frac{1}{x^3})}{(2+\frac{1}{x}-\frac{5}{x^2}+\frac{7}{x^3})}$

When $x\to \infty$, then $\frac{1}{x}\to 0$

$=\frac{(3-4(0)+6(0)-0)}{(2+0-5(0)+7(0\left)\right)}=\frac{3}{2}$

Therefore,
$\underset{x\to \infty }{lim}\frac{3x^3-4x^2+6x-1}{2x^3-x^2+5x+7}=\frac{3}{2}$