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 26E

(a)

To determine

To estimate: The value of the function to two decimal places when x approaches zero by using the graph of the function f(x).

Expert Solution

Answer to Problem 26E

The estimated value of the function when x approaches zero f(x) approaches to 0.29.

Explanation of Solution

Given:

The function f(x)=3+x3x.

Use the online graphing calculator to draw the graph of the function f(x)=3+x3x.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 2.3, Problem 26E

From the graph, as x=0, then f(x) is not defined. But x approaches 0, and then f(x) tends to 0.289.

That is, limx03+x3x0.29.

Thus, the estimated value of limx03+x3x is 0.29_.

(b)

To determine

To estimate: The value of the limit to four decimal places when x close to 0 by using the table of values of f(x).

Expert Solution

Answer to Problem 26E

The estimated value of the limit by using the table of values of f(x) for x close to 0 is 0.2887_.

Explanation of Solution

Calculation:

Construct the table of values of f(x) for x close to 0.

xf(x)=3+x3x
−0.0013+(0.001)30.0010.2886992
−0.000 13+(0.0001)30.00010.2886775
−0.000 013+(0.00001)30.000010.2886754
−0.000 0013+(0.000001)30.0000010.2886752
0.000 0013+(0.000001)30.0000010.2886751
0.000 013+(0.00001)30.000010.2886749
0.000 13+(0.0001)30.00010.2886727
0.0013+(0.001)30.0010.2886511

From the table, f(x) approaches 0.2887 as x gets more close to 0.

That is, limx03+x3x0.2887.

Thus, the limit appears to be approximately equal to 0.2887_.

(c)

To determine

To find: The exact limit value of the function f(x)=3+x3x as x approaches 0.

Expert Solution

Answer to Problem 26E

The exact limit value of the function is 123_.

Explanation of Solution

Given:

The limit of the function as x approaches 0 is f(x)=3+x3x.

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 7: limxac=c

Limit law 8: limxax=a

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

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).

Difference of square formula: (a2b2)=(a+b)(ab)

Fact 1:

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

Let f(x)=3+x3x (1)

Note 1:

The direct substitution method is not applicable for the function f(x) as the function f(0) is in indeterminate form when x=0.

f(0)=3+030=330=00

Note 2:

The Quotient rule is not applicable for the function f(x) as the limit of the denominator is zero.

limx0(x)=0[ by limit law 8]

Note 3:

The limit may be infinite or some finite value when both the numerator and the denominator approach 0.”

Calculation:

By note 3, take the limit x approaches 0 but x0.

Simplify f(x) by using elementary algebra, f(x)=3+x3x.

Take the conjugate of the numerator and multiply and divide by f(x).

f(x)=3+x3x×3+x+33+x+3=(3+x3)(3+x+3)x(3+x+3)

Apply the formula for the difference of square,

f(x)=((3+x)2(3)2)x(3+x+3)=(3+x3)x(3+x+3)=xx(3+x+3)

Since the limit x approaches 0 but not equal to 0, cancel the common term x0 from both the numerator and the denominator,

f(x)=1(3+x+3)

Use fact 1, f(x)=1(3+x+3) and x0, then limx03+x3x=limx01(3+x+3).

Use the limit laws to find the required limit function.

limx01(3+x+3)=limx01limx0(3+x+3)[by limit law 5]=limx01limx0(3+x)+limx03[by limit law 1]=1limx0(3+x)+3[by limit law 7]

=1limx0(3+x)+3[by limit law 11]=1limx03+limx0x+3[by limit law 1]=13+(0)+3[by limit laws 7 and 8]=123

Thus, the exact limit of the function is 123_.

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