Question

Solve: $\underset{x\to \infty }{lim}(\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x})$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

Answer: (B)

$0$

When dealing with limits of ratios of polynomials of the form

$\underset{x\to \infty }{lim}\frac{f\left(x\right)}{g\left(x\right)}$ or $\underset{x\to -\infty }{lim}\frac{f\left(x\right)}{g\left(x\right)}$

If the degree of the numerator is less than the degree of the denominator, the limit is $0$.

$\underset{x\to \infty }{lim}(\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x})$

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

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

$=0-0=0$