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 2.3, Problem 11E
To determine

To evaluate: The limit of the function limx5x25x+6x5.

Expert Solution

Answer to Problem 11E

The limit of the function does not exist.

Explanation of Solution

Limit Laws:

Suppose that c is a constant and the limits limxaf(x) and limxag(x) exist, then

Limit law 2: limxa[f(x)g(x)]=limxaf(x)limxag(x)

Limit law 7: limxac=c

Limit law 8: limxax=a

Reason:

The limit of the denominator is zero.

limx5(x5)=limx5(x)limx5(5) (by limit law 2)=(5)(5) (by limit law 8 and 7)=0

The limit of the numerator is 6.

limx5(x25x+6)=limx5(x2)limx5(5x)+limx5(6) (by limit law 2 and 1)=limx5(x2)5limx5(x)+limx5(6) (by limit law 3)=(5)25(5)+6 (by limit law 9,8 and 7)=6

The quotient law cannot be used as the denominator of the function tends to zero as x5.

So, limx5x25x+6x5 does not exist as the numerator of the function approaches 6 when x tends to 5 and the denominator of the function approaches 0 when x tends to 5.

Thus, the limit of the function does not exist.

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