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 5RE
To determine

To evaluate: The limit of the function limx3x29x2+2x3.

Expert Solution

Answer to Problem 5RE

The limit of the function is 32_.

Explanation of Solution

Formula used:

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.

Given:

Let f(x)=x29x2+2x3 (1)

Note 1:

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

f(3)=(3)29(3)2+2(3)3=99963=099=00

Note 2:

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

Calculation:

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

Simplify f(x) by using elementary algebra.

f(x)=x29x2+2x3=x2(3)2x2+2x3

Apply the difference of square formula in the numerator,

f(x)=(x+3)(x3)x2+2x3 (2)

Factorise the denominator of f(x).

x2+2x3=x2+3xx3=x(x+3)(x+3)=(x1)(x+3)

Substitute (x1)(x+3) for x2+2x3 in equation (1).

f(x)=(x+3)(x3)(x1)(x+3)

Since the limit of x approaches to −3 but not equal to −3, cancel the common term x+3(0) from both the numerator and the denominator,

f(x)=(x3)(x1)

Use fact 1, f(x)=(x3)(x1) and x3, then limx3x29x2+2x3=limx3(x3)(x1).

Apply the direct substitution property on the limit function.

limx3(x3)(x1)=3331=64=32

Thus, the limit of the function is 32_.

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