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Question: How do you find vertical, horizontal and oblique asymptotes for \(F\left( x \right)=\dfrac{x-1}{x-{{...

How do you find vertical, horizontal and oblique asymptotes for F(x)=x−1x−x3?F\left( x \right)=\dfrac{x-1}{x-{{x}^{3}}}?

Explanation

Solution

A line y=by=b is a horizontal asymptote of the graph of a function y=f(x)y=f\left( x \right) if either lim⁡x→∞f(x)=b\displaystyle \lim_{x \to \infty }f\left( x \right)=b or lim⁡x→−∞f(x)=b.\displaystyle \lim_{x \to -\infty }f\left( x \right)=b. A line x=ax=a is a vertical asymptote of a functiony=f(x)y=f\left( x \right) if either lim⁡x→a+f(x)=±∞\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\pm \infty or lim⁡x→a−f(x)=±∞.\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\pm \infty .

Complete step-by-step solution:
Let us consider the given function F(x)=x−1x−x3.F\left( x \right)=\dfrac{x-1}{x-{{x}^{3}}}.
We are asked to find the vertical, horizontal and oblique asymptotes for this function.
Let us simplify the given function as F(x)=x−1x−x3=x−1x(1−x2)=x−1−x(x2−1).F\left( x \right)=\dfrac{x-1}{x-{{x}^{3}}}=\dfrac{x-1}{x\left( 1-{{x}^{2}} \right)}=\dfrac{x-1}{-x\left( {{x}^{2}}-1 \right)}.
Now, we will apply the identity (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} in the denominator.
We will get F(x)=x−1−x(x+1)(x−1)=−1x(x+1)=−1x2+x.F\left( x \right)=\dfrac{x-1}{-x\left( x+1 \right)\left( x-1 \right)}=\dfrac{-1}{x\left( x+1 \right)}=\dfrac{-1}{{{x}^{2}}+x}.
Let us first discuss how to find the horizontal asymptotes for the given function.
A line y=by=b is a horizontal asymptote of the graph of a function y=f(x)y=f\left( x \right) if either lim⁡x→∞f(x)=b\displaystyle \lim_{x \to \infty }f\left( x \right)=b or lim⁡x→−∞f(x)=b.\displaystyle \lim_{x \to -\infty }f\left( x \right)=b.
We will get lim⁡x→±∞f(x)=lim⁡x→±∞−1x2+x=0.\displaystyle \lim_{x \to \pm \infty }f\left( x \right)=\displaystyle \lim_{x \to \pm \infty }\dfrac{-1}{{{x}^{2}}+x}=0.
Therefore, there is a horizontal asymptote at y=0.y=0.
Let us first discuss how to find the vertical asymptotes for the given function.
A line x=ax=a is a vertical asymptote of a functiony=f(x)y=f\left( x \right) if either lim⁡x→a+f(x)=±∞\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\pm \infty or lim⁡x→a−f(x)=±∞.\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\pm \infty .
We need to find the limit of the function at x=ax=a from the left side and the limit of the function at x=ax=a from the right side.
Now, lim⁡x→0+f(x)=lim⁡x→0+−1x2−x=∞.\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{-1}{{{x}^{2}}-x}=\infty .
And lim⁡x→0−f(x)=lim⁡x→0−−1x2−x=−∞.\displaystyle \lim_{x \to {{0}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{-1}{{{x}^{2}}-x}=-\infty .
Now, we will get x=0x=0 is a vertical asymptote.
Also, we will get lim⁡x→−1+f(x)=lim⁡x→−1+−1x2+x=0.\displaystyle \lim_{x \to -{{1}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to -{{1}^{+}}}\dfrac{-1}{{{x}^{2}}+x}=0.
And lim⁡x→−1−f(x)=lim⁡x→−1−−1x2−x=0.\displaystyle \lim_{x \to -{{1}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to -{{1}^{-}}}\dfrac{-1}{{{x}^{2}}-x}=0.
Therefore x=−1x=-1 is also a vertical asymptote.
By symmetry, the same is applicable for the line x=1.x=1.
We know that the oblique asymptote occurs when the degree of denominator is one less than the degree of numerator. Therefore, this function does not have oblique asymptotes.
Hence the given function has a horizontal asymptote at y=0,y=0, vertical asymptotes at x=0,±1x=0,\pm 1 and no oblique asymptotes.

Note: We know that a straight line is said to be an asymptote of an infinite branch of the graph of a function if, as the point PP recedes to infinity along the branch, the perpendicular distance of PP from the straight line tends to zero.