   Chapter 2.6, Problem 64E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the limits as x → ∞ and as x → –∞. Use this information, together with intercepts, to give a rough sketch of the graph as in Example 12.Example 12 FIGURE 13y = (x – 2)4 (x + 1)3 (x – 1) y = x2(x2 – 1)2(x + 2)

To determine

To find: The limit of the function y=x2(x21)2(x+2) as x approaches infinity and negative infinity and obtain the x and y intercepts and sketch the rough graph of the function. f(x)=x2(x21)2(x+2).

Explanation

Limit Laws used: Suppose that c is a constant and the limits limxaf(x) and limxag(x) exists, then

Limit law 1: limxa[f(x)+g(x)]=limxaf(x)+limxag(x)

Limit law 2: limxa[f(x)g(x)]=limxaf(x)limxag(x)

Limit law 3: limxa[cf(x)]=climxaf(x)

Theorem used:

1. If r>0 is a rational number, then limx1xr=0.

2. If r>0 is a rational number such that xr is defined, then limx1xr=0.

Note:

The limit law limxaf(x)g(x)=limxaf(x)limxag(x) is valid for one sided limits. That is, if a=±, and also for infinite limits using the rules b×= if b>0.

Calculation:

Obtain the x and y intercepts.

Consider the function, f(x)=x2(x21)2(x+2).

The y-intercept is computed as follows,

Plug x=0 in the function f(x),

f(0)=0(01)2(0+2)=0

Thus, the y-intercept is 0.

The x-intercept is computed as follows,

Plug y=0 in the function f(x),

x2(x21)2(x+2)=0x2((x+1)(x1))2(x+2)=0x2(x+1)2(x1)2(x+2)=0x=0,1,1 and 2

Thus, the x intercepts are −2, −1, 0 and 1.

Compute the value of the function as x approaches infinity.

Take the limit of f(x) as x approaches infinity.

limxx2(x21)2(x+2)=limx(x2(x21)2(x+2))=limx(x2(x21)(x21)(x+2))=limx(x7+2x6x52x4x52x4+x3+2x2)=limx(x7+2x62x54x4+x3+2x2)

=limxx7(1+2x2x24x3+1x4+2x5)

Here, x7 goes to infinity and (1+2x2x24x3+1x4+2x5) goes to 1 as x approaches infinity. Thus, the product f(x)=x7(1+2x2x24x3+1x4+2x5) approaches infinity as x approaches infinity. That is,

limxf(x)=limxx7limx(1+2x2x24x3+1x4+2x5)[ by note ]=limxx7[limx(1)+2limx(1x)2limx(1x2)4limx(1x3)+limx(1x4)+2limx(1x5)][by limit law 1,2,3]=limxx7[1+2(0)2(0)4(0)+(0)+2(0)][by theorem 1]=()

=

Thus, the function f(x)=x2(x21)2(x+2) approaches infinity as x approaches infinity

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