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 9E
To determine

To evaluate: The limit of the function limx5x26x+5x5.

Expert Solution

Answer to Problem 9E

The limit of the function is 4.

Explanation of Solution

Limit Laws:

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

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

Limit law 7: limxac=c

Limit law 8: limxax=a

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).

Fact 1:

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

Calculation:

Compute the limit value of the denominator.

limx5(x5)=limx5(x)limx5(5) (by limit law 2)=limx5(x)5 (by limit law 7)=55 (by limit law 8)=0

Since the limit of the denominator is zero, the quotient law cannot be applied.

Let f(x)=x26x+5x5 (1)

Simplify f(x) by using elementary algebra.

Factorise the numerator x26x+5.

x26x+5=x25x1x+5=x(x5)(x5)=(x5)(x1)

Substitute (x5)(x1) for x26x+5 in equation (1).

f(x)=(x5)(x1)x5

Cancel the common terms from both the numerator and the denominator, f(x)=(x1).

Use fact 1, f(x)=(x1) and x5, then limx5x26x+5x5=limx5(x1).

Apply direct substitution property on the limit function.

limx5(x1)=51=4

Thus, the limit of the function is 4.

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