(a) The mathematical proof of the statement stating that using L'Hopital Rule, f ( x ) = e − x − e − a x x , though not defined at x = 0, can be made continuous by assigning the value f ( 0 ) = a − 1

Question

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Answer 1

Question

Chapter 16.1, Problem 55E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

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Answer 2

Question

Chapter 16.1, Problem 55E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Expert Solution & Answer

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Answer 3

Question

Chapter 16.1, Problem 55E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Expert Solution & Answer

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Answer 4

Question

Chapter 16.1, Problem 55E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

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See solution

Answer 5

Chapter 16.1, Problem 55E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Answer 6

Chapter 16.1, Problem 55E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Answer 7

Chapter 16.1, Problem 55E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Answer 8

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Answer 9

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Answer 10

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Answer 11

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Answer 12

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Answer 13

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Answer 14

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Answer 15

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Answer 16

Expert Solution & Answer

Answer 17

Expert Solution & Answer

Answer 18

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Answer 19

Ch. 16.1 - Prob. 1PQ Ch. 16.1 - Prob. 2PQ Ch. 16.1 - Prob. 3PQ Ch. 16.1 - Prob. 4PQ Ch. 16.1 - Prob. 5PQ Ch. 16.1 - Prob. 6PQ Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E

(a) The mathematical proof of the statement stating that using L'Hopital Rule, f ( x ) = e − x − e − a x x , though not defined at x = 0, can be made continuous by assigning the value f ( 0 ) = a − 1

Solution Summary: The author explains how the mathematical proof of the statement stating that using L'Hopital Rule, f ( x ) = e

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(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

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Chapter 16 Solutions

(a) The mathematical proof of the statement stating that using L'Hopital Rule, f ( x ) = e − x − e − a x x , though not defined at x = 0, can be made continuous by assigning the value f ( 0 ) = a − 1

Solution Summary: The author explains how the mathematical proof of the statement stating that using L'Hopital Rule, f ( x ) = e

Videos