Question

Which of the following is/are true for $y=\frac{x^3}{x^2+1}$

• A. It has no vertical asymptote
• B. It has no horizontal asymptote
• C. It has no inclined asymptote
• D. Its inclined asymptote passes through origin

Answer 1

Answer: (A), (B), (D)

Its inclined asymptote passes through origin

Given: $y=\frac{x^3}{x^2+1},$ Domain is $R$
$\because y$ is defined $\forall x\in R,$ it has no vertical asymptote.
$\because$Degree of numerator is greater than that of denominator, it has no horizontal asymptote.
Let $y=mx+c$ be inclined asymptote (if it exists)
$m=\underset{x\to \pm \infty }{lim}\left(\frac{y}{x}\right)=\underset{x\to \pm \infty }{lim}\left(\frac{x^2}{x^2+1}\right)=1$
$c=\underset{x\to \pm \infty }{lim}(y-mx)=\underset{x\to \pm \infty }{lim}(\frac{x^3}{x^2+1}-x)=\underset{x\to \pm \infty }{lim}\left(\frac{-x}{x^2+1}\right)=0$
$\therefore$ Inclined asymptote is $y=x$ and it passes through origin.