Chapter 1: Limits
Section 1.5: Limits at Infinity and Infinite Limits
Example 1.5.6
Graph the rational function fx=2x3−3x2+5x−7x2−x−6 and determine all its asymptotes.
Solution
The solution in Table 1.5.6(a) is generated by calling the Rational Function tutor (Student Precalculus package) via a , the path to which is shown at the top of the table.
Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes
Rational Function Tutor
Enter a rational function PxQx:
Asymptotes
Horizontal
Oblique
Vertical
Plot
Table 1.5.6(a) Solution of Example 1.5.6 via the Rational Function tutor, called from a task template
Enter the rule for fx in the space above the "Asymptotes" button. Press that button to have the equations of the asymptotes appear in the spaces to the right. There are no horizontal asymptotes, but two vertical asymptotes (the lines x=−2, and x=3). In addition, the line y=2 x−1 is an oblique asymptote because the degree of the numerator is one more than the degree of the denominator.
Press the "Rational Function Tutor" button to launch the Rational Function tutor. Figure 1.5.6(a) is a screenshot of this tutor as it appears after it is launched.
If the tutor is launched from the Tools≻Tutors: Precalculus menu, the numerator and denominator of the rational function f have to be entered separately. By launching it from the task template, this data-entry chore is simplified. In the space below the numerator and denominator, the equations of the asymptotes appear. However, when this tutor is closed, all this information is lost, and only the graph is preserved.
Figure 1.5.6(a) Screen-shot of Rational Function tutor
Accessing this tutor from the task template preserves the equations of the asymptotes, and also the graph, which is written to the task template when the Close button on the tutor is pressed. The Maple command (RationalFunctionPlot) at the bottom of the tutor can be copied, then pasted and executed; its output is the graph seen in the tutor.
Alternate Solution
Initialize
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
Control-drag (or type) fx=…
Context Panel: Assign Function
fx=2x3−3x2+5x−7x2−x−6→assign as functionf
Obtain the equations of the asymptotes
Apply the Asymptotes command.
Asymptotesfx
y=2x−1,x=−2,x=3
Obtain the limits at x=±∞
Expression palette: Limit operator
limx→∞fx = ∞
limx→−∞fx = −∞
Obtain the one-sided limits at x=−2
limx→−2−fx = −∞
limx→−2+fx = ∞
Obtain the one-sided limits at x=3
limx→3−fx = −∞
limx→3+fx = ∞
Obtain the equation of the oblique asymptote
Type fx and press the Enter key.
Context Panel: Numerator
Context Panel: Assign to a Name≻p
fx
2x3−3x2+5x−7x2−x−6
→numerator
2x3−3x2+5x−7
→assign to a name
p
Context Panel: Assign to a Name≻q
→denominator
x2−x−6
q
Implement long division of p by q via the quo command. The remainder is assigned to r.
quop,q,x,'r'
2x−1
Show that pq=2 x−1+rq=fx
2 x−1+rq = 2x−1+−13+16xx2−x−6= simplify 2x3−3x2+5x−7x2−x−6
Evaluating Limits
For limits at ±∞, divide numerator and denominator by x2
Divide the numerator by x2.
px2 = 2x3−3x2+5x−7x2= expand 2x−3+5x−7x2
Divide the denominator by x2.
qx2 = x2−x−6x2= expand 1−1x−6x2
Control-drag to form p/x2q/x2. The dominant term in the numerator is 2 x, from which the limits at ±∞ readily follow.
2x−3+5x−7x21−1x−6x2
Zeros of the denominator and vertical asymptotes
Obtain the zeros of the denominator.
q = x2−x−6→solvex=3,x=−2
Context Panel: Evaluate at a Point≻x=−2
p = 2x3−3x2+5x−7→evaluate at point−45
Context Panel: Evaluate at a Point≻x=3
p = 2x3−3x2+5x−7→evaluate at point35
See Figure 1.5.6(b) for a graph of the denominator.
To the left of x=−2, the denominator is positive and the numerator is negative (approximately −45). Consequently, f→−∞ as x→−2 from the left.
To the right of x=−2, the denominator is negative and the numerator is negative (approximately −45). Consequently, f→+∞ as x→−2 from the right.
Figure 1.5.6(b) Graph of the denominator of fx
To the left of x=3, the denominator is negative and the numerator is positive (approximately 35). Consequently, f→−∞ as x→3 from the left.
To the right of x=3, the denominator is positive and the numerator is positive (approximately 35). Consequently, f→+∞ as x→3 from the right.
If the numerator is nonzero at a zero of the denominator, that zero determines the location of a vertical asymptote. What then remains is to determine the one-sided limits on either side of the vertical asymptote.
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