Question

How do you find the vertical, horizontal or slant asymptotes for $f\left( x \right)=\dfrac{{{x}^{3}}}{{{\left( x-1 \right)}^{2}}}$?

Answer 1

An asymptote for a function is defined as the line to which the graph of the function approaches as the graph approaches infinity. Since the function is given as $f\left( x \right)=\dfrac{{{x}^{3}}}{{{\left( x-1 \right)}^{2}}}$, we can set the denominator equal to zero to get the equation of the vertical asymptote. For finding the slant or the oblique asymptote, we have to divide the numerator of the function by the denominator. Then on considering the limit as x tends to infinity, we will get the equation for the slant asymptote as well.

Complete step by step solution:
The function given to us in the above question is
$\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{{{\left( x-1 \right)}^{2}}}$
We know that the asymptote is a line to which the graph of the function approaches as the graph approaches infinity. Therefore, we consider the case when the value of the function $f\left( x \right)$ approaches infinity. This can be obtained by putting the denominator of the function to zero to get
$\Rightarrow {{\left( x-1 \right)}^{2}}=0$
On solving the above equation, we get
$\Rightarrow x=1$
Thus, we got the vertical asymptote for the given function as $x=1$.
Now, to determine the slant asymptote, we have to simplify the given function by dividing the numerator by the denominator. But before that, we have to simplify the denominator for which we consider the given function
$\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{{{\left( x-1 \right)}^{2}}}$
By using the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we can simplify the denominator of the above function as
$\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{{{x}^{2}}-2x+1}$
Now, we divide the numerator ${{x}^{3}}$ by the denominator ${{x}^{2}}-2x+1$ as shown below.

& {{x}^{3}} \\\ & \underline{{{x}^{3}}-2{{x}^{2}}+x} \\\ & 2{{x}^{2}}-x \\\ & \underline{2{{x}^{2}}-4x+2} \\\ & \underline{3x-2} \\\ \end{aligned}}\right.}}$$ From the above division, we can write the given function as $$\Rightarrow f\left( x \right)=x+2+\dfrac{3x-2}{{{x}^{2}}-2x+1}$$ Considering the limit $x \to \infty $ both the sides, we get $$\begin{aligned} & \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }\left[ \left( x+2 \right)+\dfrac{3x-2}{{{x}^{2}}-2x+1} \right] \\\ & \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }\left( x+2 \right)+\displaystyle \lim_{x \to \infty }\left( \dfrac{3x-2}{{{x}^{2}}-2x+1} \right) \\\ \end{aligned}$$ Taking $x$ and ${{x}^{2}}$ common respectively from the numerator and the denominator of the second tersm, we get $$\begin{aligned} & \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }\left( x+2 \right)+\displaystyle \lim_{x \to \infty }\left( \dfrac{x\left( 3-\dfrac{2}{x} \right)}{{{x}^{2}}\left( 1-\dfrac{2}{x}+\dfrac{1}{{{x}^{2}}} \right)} \right) \\\ & \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }\left( x+2 \right)+\displaystyle \lim_{x \to \infty }\left( \dfrac{\left( 3-0 \right)}{x\left( 1-0+0 \right)} \right) \\\ & \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }\left( x+2 \right)+0 \\\ & \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }\left( x+2 \right) \\\ \end{aligned}$$ From the above equation, we can say that when x is increased infinitely, the curve approaches the straight line $y=x+2$, which is the slant asymptote for the given function. We can observe these asymptotes in the graph of the given function below.  Hence, we have obtained the asymptotes for the given function as $x=1$ and $y=x+2$. **Note:** We must not worry about the horizontal asymptote for the given function, as it does not exist. This is because the degree of the denominator in the given function is less than that of the numerator. The vertical asymptote exists only when the degree of the denominator is greater than or equal to that of the numerator.