Chapter 1: Limits
Section 1.5: Limits at Infinity and Infinite Limits
Example 1.5.4
Evaluate limx→∞sinxgx, where gx=x−1x2+x+1.
Solution
Figure 1.5.4(a) has a graph of sinxx−1x2+x+1 in red, and graphs of ±x−1x2+x+1 in blue and green, respectively. The blue and green curves are envelopes for the decreasing oscillations of sinx. The figure suggests that limx→∞sinxgx=0.
Unfortunately, since limx→∞sinx does not exist, the Product rule for limits cannot be applied. Indeed, since sinx≤1, Principle 1.1.1 could be invoked. Because limx→∞gx=0, it immediately follows that limx→∞sinxgx=0.
g1 := (x-1)/(x^2+x+1): f1 := sin(x)*g1: plot([f1,abs(g1),-abs(g1)],x=10..200,color=[red,blue,green]);
Figure 1.5.4(a) Graph of sinxx−1x2+x+1
Application of Maple's limit operator
Expression palette: Limit operator A space or explicit multiplication between factors in the numerator is essential.
Context Panel: Evaluate and Display Inline
limx→∞sinx x−1x2+x+1 = 0
Application of the Squeeze theorem
With gx=x−1x2+x+1, and sinxgx≤gx because sinx≤1, it follows that
−gx≤sinxgx≤gx
or
−x−1x2+x+1≤sinxx−1x2+x+1≤x−1x2+x+1
Since limx→∞gx=limx→∞−gx=0, both sides of the inequality go to zero, so the middle term must also go to zero, by the Squeeze theorem.
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