Question

What are the asymptotes of $f\left( x \right)=\dfrac{2x-1}{x-2}$?

Answer 1

This type of question depends on the concept of asymptotes of rational function. A rational function has vertical and horizontal asymptotes. A vertical asymptote occurs where f(x) is undefined that is denominator equals zero and hence to find, vertical asymptote we have to equate denominator to zero. Horizontal asymptotes are found by comparing the degree of the denominator to the degree of the numerator. If the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y=0$. If the degree of numerator is equal to the degree of denominator then horizontal asymptote occur as $\displaystyle \lim_{x \to \pm \infty }f\left( x \right)\to c,c\text{ is constant}$ and hence to find the horizontal asymptote we have to find $\displaystyle \lim_{x \to \pm \infty }f\left( x \right)$

Complete step by step answer:
Now, we have to find the vertical and horizontal asymptotes for $f\left( x \right)=\dfrac{2x-1}{x-2}$
By equating denominator to zero we can find vertical asymptote, so consider,
$\Rightarrow \left( x-2 \right)=0$
On solving we get,
$\Rightarrow x=2$ is the vertical asymptote.
Now, horizontal asymptotes occur as $\displaystyle \lim_{x \to \pm \infty }f\left( x \right)\to c,c\text{ is constant}$. So, consider,
$\Rightarrow \displaystyle \lim_{x \to \pm \infty }f\left( x \right)=\displaystyle \lim_{x \to \pm \infty }\left( \dfrac{2x-1}{x-2} \right)$
By dividing numerator as well as denominator by ‘x’, we get,
$\Rightarrow \displaystyle \lim_{x \to \pm \infty }f\left( x \right)=\displaystyle \lim_{x \to \pm \infty }\left( \dfrac{\dfrac{2x}{x}-\dfrac{1}{x}}{\dfrac{x}{x}-\dfrac{2}{x}} \right)=\displaystyle \lim_{x \to \pm \infty }\left( \dfrac{2-\dfrac{1}{x}}{1-\dfrac{2}{x}} \right)$
As $x \to \pm \infty ,f\left( x \right)\to \dfrac{2-0}{1-0}$
$\Rightarrow y=f\left( x \right)=2$ is the horizontal asymptote.
Hence, the function $f\left( x \right)=\dfrac{2x-1}{x-2}$ has vertical asymptote at $x=2$ and horizontal asymptote at $y=2$.

Note: In this type of question students may make mistakes in calculation of limits. Also when we find a vertical asymptote at that time we have to take care that the value of the variable does not result in a zero numerator. That means after obtaining a vertical asymptote we have to check the value of the numerator, it must be non-zero.