# Whether the statement, lim x → 1 ( x 2 + 6 x − 7 x 2 + 5 x − 6 ) = lim x → 1 ( x 2 + 6 x − 7 ) lim x → 1 ( x 2 + 5 x − 6 ) is true or false. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 2, Problem 2RQ
To determine

## Whether the statement, limx→1(x2+6x−7x2+5x−6)=limx→1(x2+6x−7)limx→1(x2+5x−6) is true or false.

Expert Solution

The statement is false.

### Explanation of Solution

Quotient Law: Suppose that the limits limxaf(x) and limxag(x) exist.

Then limxaf(x)g(x)=limxaf(x)limxag(x) if limxag(x)0

Reason:

Let the function Q(x)=f(x)g(x) where f(x)=x2+6x7 and g(x)=x2+5x6.

Then, by the Quotient Law, the given statement is true only if the limit of the denominator Q(x) is not equal to zero as x approaches 1.

That is, limx1(x2+6x7x2+5x6)=limx1(x2+6x7)limx1(x2+5x6) is true whenever limx1(x2+5x6) is not equal to zero.

The limit g(x)=x2+5x6 as x approaches 1 is computed as follows,

limx1(x2+5x6)=(1)2+5(1)6=1+56=66=0

Thus, the limit exists and it is equal to zero.

Since the denominator g(x)=x2+5x6 is equal to zero as x approaches 1, the Quotient law cannot be used here.

Therefore, the statement is false.

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