BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2, Problem 12RE
To determine

To evaluate: The limit of the function limx3x+6xx33x2.

Expert Solution

Answer to Problem 12RE

The limit of the function is 554_.

Explanation of Solution

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then limxaf(x)=f(a).

Difference of squared formula: (a2b2)=(a+b)(ab)

Fact 1:

If f(x)=g(x) when xa, then limxaf(x)=limxag(x), provided the limit exist.

Evaluation:

Let f(x)=x+6xx33x2 (1)

Note 1:

The direct substitution method is not applicable for the function f(x) as the function f(3) is in an indeterminate form when x=3.

f(3)=3+63(3)33(3)2=93273(9)=332727=00

Note 2:

The limit may be infinite or some finite value when both the numerator and the denominator approach 0.”

Calculation:

By note 2, consider the limit x approaches to 3 but x3.

Simplify f(x) by using elementary algebra.

f(x)=x+6xx33x2=x+6xx2(x3)

Multiply both the numerator and the denominator by the conjugate of the numerator,

f(x)=x+6xx33x2×x+6+xx+6+x

Apply the difference of squared formula,

f(x)=(x+6)2(x)2x2(x3)(x+6+x)=x+6x2x2(x3)(x+6+x)

=(x2x6)x2(x3)(x+6+x) (2)

Factorize the numerator,

x2x6=x23x+2x6=x(x3)+2(x3)=(x3)(x+2)

Substitute (x3)(x+2) for (x2x6) in equation (2),

f(x)=(x3)(x+2)x2(x3)(x+6+x)

Since the limit of x approaches 3 but not equal to 3, cancel the common term x30 from both the numerator and the denominator,

f(x)=(x+2)x2(x+6+x).

Use fact 1, f(x)=(x+2)x2(x+6+x) and x3, then limx3x+6xx33x2=limx3(x+2)x2(x+6+x).

Apply the direct substitution property on the limit function.

limx3(x+2)x2(x+6+x)=(3+2)(3)2(3+6+3)=(5)9(9+3)=(5)9(3+3)=554

Thus, the limit of the function is 554_.

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