Asymptotes
Solved Problems
Example 5.
Find the asymptotes of the function \[f\left( x \right) = \frac{{3{x^2} - 2x + 1}}{{x - 1}}.\]
Solution.
It is clear that the line \(x = 1\) is a vertical asymptote, because at this point the function has a discontinuity and the following conditions are true:
\[\lim\limits_{x \to 1 - 0} f\left( x \right) = \lim\limits_{x \to 1 - 0} \frac{{3{x^2} - 2x + 1}}{{x - 1}} = \frac{{3{{\left( {1 - 0} \right)}^2} - 2\left( {1 - 0} \right) + 1}}{{1 - 0 - 1}} = - \infty ,\]
\[\lim\limits_{x \to 1 + 0} f\left( x \right) = \lim\limits_{x \to 1 + 0} \frac{{3{x^2} - 2x + 1}}{{x - 1}} = \frac{{3{{\left( {1 + 0} \right)}^2} - 2\left( {1 + 0} \right) + 1}}{{1 + 0 - 1}} = + \infty .\]
We write the function as
\[y=f\left( x \right) = \frac{{3{x^2} - 2x + 1}}{{x - 1}} = \frac{{3{x^2} - 3x + x - 1 + 2}}{{x - 1}} = \frac{{3x\cancel{\left( {x - 1} \right)}}}{\cancel{x - 1}} + \frac{\cancel{x - 1}}{\cancel{x - 1}} + \frac{2}{{x - 1}} = 3x + 1 + \frac{2}{{x - 1}} = 3x + 1 + \alpha \left( x \right),\]
where \(\alpha \left( x \right) \to 0\) as \(x \to \pm \infty.\)
Thus, the function has the oblique asymptote \(y = 3x + 1.\)
Note that a rational function may have an oblique asymptote if the degree of the numerator is one greater than the degree of the denominator. A schematic view of this curve is shown in Figure \(8.\)
Figure 8.
Example 6.
Find the asymptotes of the function \[f\left( x \right) = {\frac{{{x^2}}}{{x - 1}}}.\]
Solution.
The function is defined for all \(x \in \mathbb{R}\) except \(x = 1\) where the denominator is equal zero. Look for vertical asymptotes:
\[\lim\limits_{x \to 1 - 0} f\left( x \right) = \lim\limits_{x \to 1 - 0} \frac{{{x^2}}}{{x - 1}} = - \infty ;\]
\[\lim\limits_{x \to 1 + 0} f\left( x \right) = \lim\limits_{x \to 1 + 0} \frac{{{x^2}}}{{x - 1}} = + \infty .\]
Hence there is a vertical asymptote given by the equation \(x = 1.\)
Notice that the degree of the numerator is one more than the degree of the denominator. Therefore, the function has an oblique asymptote. We can find its equation by long division:
\[\frac{{{x^2}}}{{x - 1}} = \frac{{{x^2} - x + x - 1 + 1}}{{x - 1}} = \frac{{x\left( {x - 1} \right) + \left( {x - 1} \right) + 1}}{{x - 1}} = \frac{{x\cancel{\left( {x - 1} \right)}}}{\cancel{x - 1}} + \frac{\cancel{x - 1}}{\cancel{x - 1}} + \frac{1}{{x - 1}} = x + 1 + \frac{1}{{x - 1}}.\]
As \(\lim\limits_{x \to \pm \infty } \frac{1}{{x - 1}} = 0,\) the equation of the oblique asymptote is written in the form
\[y = x + 1.\]
This oblique asymptote exists for \(x \to \pm\infty.\) Hence, the function does not have horizontal asymptotes.
Figure 9.
Example 7.
Find the asymptotes of the function \[f\left( x \right) = \sqrt {x + 1} - \sqrt {x - 1} .\]
Solution.
The function is defined for \(x \ge 1.\) Calculate the following limit:
\[\lim\limits_{x \to 1 + 0} f\left( x \right) = \lim\limits_{x \to 1 + 0} \left( {\sqrt {x + 1} - \sqrt {x - 1} } \right) = \sqrt 2 - \sqrt 0 = \sqrt 2 .\]
As this limit is finite, the function has no vertical asymptote.
Check for a horizontal asymptote:
\[\lim\limits_{x \to + \infty } f\left( x \right) = \lim\limits_{x \to + \infty } \left( {\sqrt {x + 1} - \sqrt {x - 1} } \right) = \left[ {\infty - \infty } \right] = \lim\limits_{x \to + \infty } \frac{{{{\left( {\sqrt {x + 1} } \right)}^2} - {{\left( {\sqrt {x - 1} } \right)}^2}}}{{\sqrt {x + 1} + \sqrt {x - 1} }} = \lim\limits_{x \to + \infty } \frac{{\cancel{x} + 1 - \cancel{x} + 1}}{{\sqrt {x + 1} + \sqrt {x - 1} }} = \lim\limits_{x \to + \infty } \frac{2}{{\sqrt {x + 1} + \sqrt {x - 1} }} = 0.\]
Hence, there is a horizontal asymptote at \(y = 0.\)
The function has no oblique asymptote. Indeed, trying to calculate the slope \(k,\) we get \(k = 0\) which corresponds to the horizontal asymptote \(y = 0\) found above.
Example 8.
Find the asymptotes of the function \[f\left( x \right) = {\frac{{{x^3}}}{{{x^2} + 1}}}.\]
Solution.
The function is defined over all \(x \in \mathbb{R}\) and has no vertical asymptotes.
Check for horizontal asymptotes:
\[\lim\limits_{x \to + \infty } f\left( x \right) = \lim\limits_{x \to + \infty } \frac{{{x^3}}}{{{x^2} + 1}} = \lim \limits_{x \to + \infty } \frac{{\frac{{{x^3}}}{{{x^3}}}}}{{\frac{{{x^2}}}{{{x^3}}} + \frac{1}{{{x^3}}}}} = \lim\limits_{x \to + \infty } \frac{1}{{\frac{1}{x} + \frac{1}{{{x^3}}}}} = + \infty ;\]
\[\lim\limits_{x \to - \infty } f\left( x \right) = \lim\limits_{x \to - \infty } \frac{{{x^3}}}{{{x^2} + 1}} = \lim \limits_{x \to - \infty } \frac{{\frac{{{x^3}}}{{{x^3}}}}}{{\frac{{{x^2}}}{{{x^3}}} + \frac{1}{{{x^3}}}}} = \lim\limits_{x \to - \infty } \frac{1}{{\frac{1}{x} + \frac{1}{{{x^3}}}}} = - \infty .\]
Hence, the function has no horizontal asymptotes.
There is an oblique asymptote since the degree of the numerator (\(2\)nd) is exactly one greater than the degree of the denominator (\(1\)st). We write the rational function as follows:
\[f\left( x \right) = \frac{{{x^3}}}{{{x^2} + 1}} = \frac{{{x^3} + x - x}}{{{x^2} + 1}} = \frac{{x\left( {{x^2} + 1} \right) - x}}{{{x^2} + 1}} = x - \frac{x}{{{x^2} + 1}}.\]
Clearly that \(\lim\limits_{x \to \pm \infty } \frac{x}{{{x^2} + 1}} = 0.\) Then the oblique asymptote is given by the equation \(y = x.\)
Figure 10.
Example 9.
Find the asymptotes of the function \[f\left( x \right) = {\frac{{{x^2} - 3x + 2}}{{{x^2} + 3x + 2}}}.\]
Solution.
First we factor the numerator and denominator:
\[f\left( x \right) = \frac{{{x^2} - 3x + 2}}{{{x^2} + 3x + 2}} = \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}}.\]
Compute the limits at the singular points \(x = -1\) and \(x = -2:\)
\[\lim\limits_{x \to - 1 - 0} f\left( x \right) = \lim\limits_{x \to - 1 - 0} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{{\left( { - 2} \right) \cdot \left( { - 3} \right)}}{{\left( { - 0} \right) \cdot 1}} = \frac{6}{{ - 0}} = - \infty ;\]
\[\lim\limits_{x \to - 1 + 0} f\left( x \right) = \lim\limits_{x \to - 1 + 0} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{{\left( { - 2} \right) \cdot \left( { - 3} \right)}}{{\left( { + 0} \right) \cdot 1}} = \frac{6}{{ + 0}} = + \infty ;\]
\[\lim\limits_{x \to - 2 - 0} f\left( x \right) = \lim\limits_{x \to - 2 - 0} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{{\left( { - 3} \right) \cdot \left( { - 4} \right)}}{{\left( { - 1} \right) \cdot \left( { - 0} \right)}} = -\frac{12}{{ - 0}} = + \infty ;\]
\[\lim\limits_{x \to - 2 + 0} f\left( x \right) = \lim\limits_{x \to - 2 + 0} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{{\left( { - 3} \right) \cdot \left( { - 4} \right)}}{{\left( { - 1} \right) \cdot \left( { + 0} \right)}} = -\frac{12}{{ + 0}} = - \infty .\]
We see that the lines \(x = -1,\) \(x = -2\) are two vertical asymptotes.
Find the limits at infinities:
\[\lim\limits_{x \to \pm \infty } f\left( x \right) = \lim\limits_{x \to \pm \infty } \frac{{{x^2} - 3x + 2}}{{{x^2} + 3x + 2}} = \lim\limits_{x \to \pm \infty } \frac{{1 - \frac{3}{x} + \frac{2}{{{x^2}}}}}{{1 + \frac{3}{x} + \frac{2}{{{x^2}}}}} = 1.\]
Then the line \(y = 1\) is a horizontal asymptote of \(f\left( x \right).\)
Now let's examine the oblique asymptotes:
\[k = \lim\limits_{x \to \pm \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{{x^2} - 3x + 2}}{{x\left( {{x^2} + 3x + 2} \right)}} = \lim\limits_{x \to \pm \infty } \frac{{{x^2} - 3x + 2}}{{{x^3} + 3{x^2} + 2x}} = \lim\limits_{x \to \pm \infty } \frac{{\frac{{{x^2}}}{{{x^3}}} - \frac{{3x}}{{{x^3}}} + \frac{2}{{{x^3}}}}}{{\frac{{{x^3}}}{{{x^3}}} + \frac{{3{x^2}}}{{{x^3}}} + \frac{{2x}}{{{x^3}}}}} = \lim\limits_{x \to \pm \infty } \frac{{\frac{1}{x} - \frac{3}{{{x^2}}} + \frac{2}{{{x^3}}}}}{{1 + \frac{3}{x} + \frac{2}{{{x^2}}}}} = \frac{0}{1} = 0.\]
Hence, there is no oblique asymptote.
The graph of the function is shown schematically in Figure \(11.\)
Figure 11.
Example 10.
Find the asymptotes of the function \[f\left( x \right) = x + \arctan x.\]
Solution.
This function is continuous on the whole set of real numbers. Therefore, it does not have vertical asymptotes. Examine the oblique asymptotes:
\[k = \lim\limits_{x \to \pm \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{x + \arctan x}}{x} = \lim\limits_{x \to \pm \infty } \left( {1 + \frac{{\arctan x}}{x}} \right) = 1 + \lim\limits_{x \to \pm \infty } \frac{{\arctan x}}{x}.\]
The inverse tangent is bounded in the interval \(\left( { - {\frac{\pi }{2}}, {\frac{\pi }{2}}} \right)\). Therefore, the limit in the last expression as \(x \to \pm \infty\) is equal to zero. Hence, \(k = 1,\) and this value is the same as \(x\) approaches positive or negative infinity. Calculate the coeficient \(b\) separately for the case of \(x \to -\infty\) and \(x \to +\infty:\)
\[{b_1} = \lim\limits_{x \to - \infty } \left[ {f\left( x \right) - kx} \right] = \lim\limits_{x \to - \infty } \left[ {\cancel{x} + \arctan x - \cancel{x}} \right] = \lim\limits_{x \to - \infty } \arctan x = - \frac{\pi }{2},\]
\[{b_2} = \lim\limits_{x \to + \infty } \left[ {f\left( x \right) - kx} \right] = \lim\limits_{x \to + \infty } \arctan x = + \frac{\pi }{2}.\]
Thus, we have found two oblique asymptotes (one for the case \(x \to -\infty\) and the other for \(x \to +\infty\)). Their equations have the form:
\[x \to - \infty :\;\;y = kx + {b_1} = x - \frac{\pi }{2},\]
\[x \to + \infty :\;\;y = kx + {b_2} = x + \frac{\pi }{2}.\]
A schematic view of the function and its asymptotes is presented in Figure \(12.\)
Figure 12.
Example 11.
Find the asymptotes of the hyperbolic tangent \[f\left( x \right) = \tanh x.\]
Solution.
By definition,
\[f\left( x \right) = \tanh x = \frac{{\sinh x}}{{\cosh x}} = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}.\]
The function is defined for all \(x \in \mathbb{R}\) and has no vertical asymptotes.
Look for horizontal asymptotes:
\[\lim\limits_{x \to + \infty } f\left( x \right) = \lim\limits_{x \to + \infty } \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} = \lim\limits_{x \to + \infty } \frac{{\frac{{{e^x}}}{{{e^x}}} - \frac{{{e^{ - x}}}}{{{e^x}}}}}{{\frac{{{e^x}}}{{{e^x}}} + \frac{{{e^{ - x}}}}{{{e^x}}}}} = \lim\limits_{x \to + \infty } \frac{{1 - {e^{ - 2x}}}}{{1 + {e^{ - 2x}}}} = \frac{{1 - 0}}{{1 + 0}} = 1;\]
\[\lim\limits_{x \to - \infty } f\left( x \right) = \lim\limits_{x \to - \infty } \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} = \lim\limits_{x \to - \infty } \frac{{\frac{{{e^x}}}{{{e^{ - x}}}} - \frac{{{e^{ - x}}}}{{{e^{ - x}}}}}}{{\frac{{{e^x}}}{{{e^{ - x}}}} + \frac{{{e^{ - x}}}}{{{e^{ - x}}}}}} = \lim\limits_{x \to - \infty } \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}} = \frac{{0 - 1}}{{0 + 1}} = - 1.\]
This, the function has two horizontal asymptotes: \(y = 1\) (as \(x \to +\infty)\) and \(y = -1\) (as \(x \to -\infty).\)
Note that a function can have either a horizontal asymptote or an oblique asymptote in one direction (that is either as \(x \to \infty\) or as \(x \to -\infty),\) but not both. Clearly, that the given function does not have an oblique asymptote.
The graph of the function is sketched in Figure \(13.\)
Figure 13.
Example 12.
Find the asymptotes of the function \[f\left( x \right) = {e^{\frac{1}{x+1}}}.\]
Solution.
The function is defined for all \(x\) except \(x = -1\) where there is a discontinuity. Check for vertical asymptotes at \(x = -1:\)
\[\lim\limits_{x \to - 1 - 0} f\left( x \right) = \lim\limits_{x \to - 1 - 0} {e^{\frac{1}{{x + 1}}}} = {e^{\lim \limits_{x \to - 1 - 0} \frac{1}{{x + 1}}}} = {e^{ - \infty }} = 0;\]
\[\lim\limits_{x \to - 1 + 0} f\left( x \right) = \lim\limits_{x \to - 1 + 0} {e^{\frac{1}{{x + 1}}}} = {e^{\lim \limits_{x \to - 1 + 0} \frac{1}{{x + 1}}}} = {e^{ + \infty }} = \infty .\]
Hence, \(x =-1\) is a vertical asymptote.
Find horizontal asymptotes(s):
\[\lim\limits_{x \to + \infty } f\left( x \right) = \lim\limits_{x \to + \infty } {e^{\frac{1}{{x + 1}}}} = {e^0} = 1;\]
\[\lim\limits_{x \to - \infty } f\left( x \right) = \lim\limits_{x \to - \infty } {e^{\frac{1}{{x + 1}}}} = {e^0} = 1.\]
So \(y = 1\) is a horizontal asymptote.
The function has no oblique asymptote (since it has a two-directional horizontal asymptote).
Figure 14.
Example 13.
Find the asymptotes of the function \[f\left( x \right) = {e^{\frac{1}{x}}} + x.\]
Solution.
The function is defined for all \(x\) except the point \(x = 0\) where it has a discontinuity.
Check for vertical asymptotes at \(x = 0:\)
\[\lim\limits_{x \to 0 - 0} f\left( x \right) = \lim\limits_{x \to 0 - 0} \left( {{e^{\frac{1}{x}}} + x} \right) = {e^{ - \infty }} + 0 = 0;\]
\[\lim\limits_{x \to 0 + 0} f\left( x \right) = \lim\limits_{x \to 0 + 0} \left( {{e^{\frac{1}{x}}} + x} \right) = {e^{ + \infty }} + 0 = + \infty .\]
So \(x = 0\) is a vertical asymptote.
Look now for horizontal asymptotes:
\[\lim\limits_{x \to + \infty } f\left( x \right) = \lim\limits_{x \to + \infty } \left( {{e^{\frac{1}{x}}} + x} \right) = 0 + \infty = + \infty ;\]
\[\lim\limits_{x \to - \infty } f\left( x \right) = \lim\limits_{x \to - \infty } \left( {{e^{\frac{1}{x}}} + x} \right) = 0 - \infty = - \infty .\]
Hence, there are no horizontal asymptotes.
Finally, we examine the oblique asymptotes:
\[k = \lim\limits_{x \to + \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to + \infty } \frac{{{e^{\frac{1}{x}}} + x}}{x} = \lim\limits_{x \to + \infty } \frac{{{e^{\frac{1}{x}}}}}{x} + 1 = 0 + 1 = 1;\]
\[b = \lim\limits_{x \to + \infty } \left[ {f\left( x \right) - kx} \right]= \lim\limits_{x \to + \infty } \left( {{e^{\frac{1}{x}}} + \cancel{x} - \cancel{x}} \right) = \lim\limits_{x \to + \infty } {e^{\frac{1}{x}}} = {e^0} = 1.\]
Similarly for \(x \to -\infty.\)
Thus, the function has the oblique asymptote \(y = x+1.\) In total, it has two asymptotes: \(x = 0\) and \(y = x + 1.\)
Figure 15.
Example 14.
Find the asymptotes of the function \[f\left( x \right) = \frac{{\sqrt {1 + {x^2}} }}{{x - 1}}.\]
Solution.
The function has a discontinuity of the second kind at \(x = 1.\) Since
\[\lim\limits_{x \to 1 - 0} f\left( x \right) = \lim\limits_{x \to 1 - 0} \frac{{\sqrt {1 + {x^2}} }}{{x - 1}} = \frac{{\sqrt {1 + {{\left( {1 - 0} \right)}^2}} }}{{\left( {1 - 0} \right) - 1}} = \frac{{\sqrt 2 }}{{ - 0}} = - \infty ,\]
\[\lim\limits_{x \to 1 + 0} f\left( x \right) = \lim\limits_{x \to 1 + 0} \frac{{\sqrt {1 + {x^2}} }}{{x - 1}} = \frac{{\sqrt {1 + {{\left( {1 + 0} \right)}^2}} }}{{\left( {1 + 0} \right) - 1}} = \frac{{\sqrt 2 }}{{ + 0}} = + \infty,\]
the straight line \(x = 1\) is a vertical asymptote.
Since the growth rate of the numerator and denominator is the same, the graph also has a horizontal asymptote. As \(x \to +\infty,\) we obtain:
\[\lim\limits_{x \to + \infty } f\left( x \right) = \lim\limits_{x \to + \infty } \frac{{\sqrt {1 + {x^2}} }}{{x - 1}} = \lim\limits_{x \to + \infty } \sqrt {\frac{{1 + {x^2}}}{{{{\left( {x - 1} \right)}^2}}}} = \lim\limits_{x \to + \infty } \sqrt {\frac{{1 + {x^2}}}{{{x^2} - 2x + 1}}} = \lim\limits_{x \to + \infty } \sqrt {\frac{{\frac{1}{{{x^2}}} + 1}}{{1 - \frac{2}{x} + \frac{1}{{{x^2}}}}}} = \frac{1}{1} = 1.\]
Similarly, the limit of the function as \(x \to -\infty\) is
\[\lim\limits_{x \to - \infty } f\left( x \right) = \lim\limits_{x \to - \infty } \frac{{\sqrt {1 + {x^2}} }}{{x - 1}} = \lim\limits_{x \to - \infty } \frac{{\sqrt {1 + {x^2}} }}{{\left( { - \sqrt {{{\left( {x - 1} \right)}^2}} } \right)}} = - \lim\limits_{x \to - \infty } \sqrt {\frac{{1 + {x^2}}}{{{x^2} - 2x + 1}}} = - \lim\limits_{x \to - \infty } \sqrt {\frac{{\frac{1}{{{x^2}}} + 1}}{{1 - \frac{2}{x} + \frac{1}{{{x^2}}}}}} = - \frac{1}{1} = - 1.\]
Thus, the graph of the function has the horizontal asymptote \(y = 1\) as \(x \to +\infty\) and the asymptote \(y = -1\) as \(x \to -\infty.\) The graph of the function is sketched in Figure \(8.\)
Figure 16.
