   Chapter 2.5, Problem 15E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval. f ( x ) = x + x − 1 ,   [ 4 , ∞ ]

To determine

To show: The function f(x)=x+x4 is continuous on the interval [4,).

Explanation

Definitions used:

1. “A function f is continuous at a number a if limxaf(x)=f(a)”.

2. A function f is continuous from the right at a number a if limxa+f(x)=f(a)

3. “A function f is continuous on an interval if it is continuous at every number in the interval (If f is defined only on one side of an endpoint of the interval means continuous from the right or continuous from the left)”.

Limit Laws:

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

Limit law 1: limxa[f(x)+g(x)]=limxaf(x)+limxag(x)

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

Limit law 3: limxa[cf(x)]=climxaf(x)

Limit law 4: limxa[f(x)g(x)]=limxaf(x)limxag(x)

Limit law 5: limxaf(x)g(x)=limxaf(x)limxag(x) if limxag(x)0

Limit law 6: limxa[f(x)]n=[limxaf(x)]n where n is a positive integer.

Limit law 7: limxac=c

Limit law 8: limxax=a

Limit law 9: limxaxn=an where n is a positive integer.

Limit law 10: limxaxn=an where n is a positive integer [If n is even, assume that a>0].

Limit law 11: limxaf(x)n=limxaf(x)n where n is a positive integer. [If n is even, assume that limxaf(x)>0].

Proof:

By the definition of continuous, f is continuous at a number a>4 if limxaf(x)=f(a).

So it is enough to show that limxaf(x)=f(a) by using the limit laws.

limxaf(x)=limxa(x+x4)=limxa(x)+limxax4[by limit law 1]=limxa(x)+limxa(x4)[by limit law 11]=limxa(x)+limxaxlimxa4[by limit law 2]

Apply the appropriate laws and simplify further

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