   Chapter 2.6, Problem 29E

Chapter
Section
Textbook Problem

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

To determine

To find: The value of limx(x2+axx2+bx).

Explanation

Result used:

Limit Laws

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 Limit law 8: limxax=a

Limit law 9: limxaxn=an where n is a positive integer.

Limit law 10: limxaxn=an where n is a positive integer, if n is even, assume that a>0.

Limit law 11: limxaf(x)n=limxaf(x)n where n is a positive integer, if n is even, assume that limxaf(x)>0.

Theorem used:

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

Calculation:

Obtain the value of the function as x approaches infinity as follows.

Simplify the function f(x)=x2+axx2+bx as follows.

f(x)=(x2+axx2+bx)=(x2+axx2+bx)×(x2+ax+x2+bx)(x2+ax+x2+bx)=(x2+ax)2(x2+bx)2(x2+ax+x2+bx)=x2+axx2bx(x2+ax+x2+bx)

=axbx(x2+ax+x2+bx)

Divide both the numerator and the denominator by the highest power of x in the denominator. That is, x0

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