# To state: The Sum law ### 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 3RCC

(a)

To determine

## To state: The Sum law

Expert Solution

### Explanation of Solution

If the limits are limxaf(x) and limxag(x) exists, then the sum law gives

limxa[f(x)+g(x)]=limxaf(x)+limxag(x)

Clearly, if limxaf(x)=L and limxag(x)=M then the result is

limxa[f(x)+g(x)]=limxaf(x)+limxag(x)limxa[f(x)+g(x)]=L+M

(b)

To determine

Expert Solution

### Explanation of Solution

If the limits are limxaf(x) and limxag(x) exists, then the Difference law gives

limxa[f(x)g(x)]=limxaf(x)limxag(x)

Clearly, if limxaf(x)=L and limxag(x)=M then the result is

limxa[f(x)g(x)]=limxaf(x)limxag(x)limxa[f(x)g(x)]=LM

(c)

To determine

Expert Solution

### Explanation of Solution

If the limits are limxaf(x) exists, then the Constant multiple law gives

limxac[f(x)]=climxaf(x)

Clearly, if limxaf(x)=L then the result is

limxac[f(x)]=climxaf(x)limxac[f(x)]=c(L)

Here, c is some constant.

(d)

To determine

Expert Solution

### Explanation of Solution

If the limits are limxaf(x) and limxag(x) exists, then the product law gives

limxa[f(x)g(x)]=limxaf(x)limxag(x)

Clearly, if limxaf(x)=L and limxag(x)=M then the result is

limxa[f(x)g(x)]=limxaf(x)limxag(x)limxa[f(x)g(x)]=LM

(e)

To determine

Expert Solution

### Explanation of Solution

If the limits are limxaf(x) and limxag(x) exists, then the Quotient law gives

limxa[f(x)g(x)]=limxaf(x)limxag(x) where limxag(x)0

Clearly, if limxaf(x)=L and limxag(x)=M then the result is

limxa[f(x)g(x)]=limxaf(x)limxag(x)limxa[f(x)g(x)]=LMM0

(f)

To determine

Expert Solution

### Explanation of Solution

If n is an integer, and the limxaf(x) exists, then the Power law gives

limxaf(x)n=(limxaf(x))n

Clearly, if limxaf(x)=L then

limxaf(x)n=(limxaf(x))nlimxaf(x)n=Ln.

(g)

To determine

Expert Solution

### Explanation of Solution

If n is an integer, and the limxaf(x) exists, and the limit is positive if n is even, then

limxaf(x)n=limxaf(x)n

Clearly, if limxaf(x)=L then the Root law gives

limxaf(x)n=limxaf(x)nlimxaf(x)n=Ln.

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