Question

The asymptotes of the function $f\left(x\right)=\frac{x^2+1}{x-1}$ is/are:

• A. $x=1$
• B. $y=1$
• C. $y=x+1.$
• D. $y=2x+1.$

Answer 1

Answer: (A), (C)

$y=x+1.$

The function is defined for all $x$ except the point $x=1$ where it has a discontinuity.

Look for vertical asymptote near $x=1$

$\underset{x\to 1–}{lim}f\left(x\right)=\underset{x\to 1–}{lim}\frac{x^2+1}{x–1}=–\infty ;$

$\underset{x\to 1+}{lim}f\left(x\right)=\underset{x\to 1+}{lim}\frac{x^2+1}{x–1}=\infty .$
There is a vertical asymptote at $x=1.$
Now, rewriting the function in the form:

$f\left(x\right)=\frac{x^2+1}{x–1}=\frac{x^2–x+x–1+2}{x–1}=\frac{x\left(x–1\right)+x–1+2}{x–1}=x+1+\frac{2}{x–1}$
where $\frac{2}{x-1}\to 0$ as $x\to \infty .$
Hence the function has an oblique asymptote at $y=x+1$