   Chapter 2.6, Problem 21E

Chapter
Section
Textbook Problem

Find the limit or show that it does not exist. lim x → ∞ ( 2 x 2 − 1 ) 2 ( x − 1 ) 2 ( x 2 + x )

To determine

To find: The value of limx(2x2+1)2(x1)2(x2+x).

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)

Limit law 4: limxa[f(x)g(x)]=limxaf(x)limxag(x)

Limit law 5: limxaf(x)g(x)=limxaf(x)limxag(x) if limxag(x)0

Limit law 6: limxa[f(x)]n=[limxaf(x)]n where n is a positive integer.

Limit law 7: limxac=c

Theorem used: If r>0 is a rational number, then limx1xr=0.

Calculation:

Obtain the value of the function as x approaches infinity.

Consider the function f(x)=(2x2+1)2(x1)2(x2+x).

Expand the function f(x) as follows,

f(x)=(2x2)2+(1)2+22x2(x2+(1)22x)(x2+x)=4x4+1+4x2(x22x+1)(x2+x)=(4x4+4x2+1)(x4+x32x32x2+x2+x)=(4x4+4x2+1)(x4x3x2+x)

Divide both the numerator and the denominator by the highest power of x in the denominator. That is, x40.

f(x)=(4x4+4x2+1)x4(x4x3x2+x)x4=(4x4x4+4x2x4+1x4)(x4x4x3x4x2x4+xx4)=4+4x2+1x411x1x2+1x3

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

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