Explanation Given: a > 0 , n > 1 , f ( x ) = x n e a x − 1 for x ≠ 0 and f ( 0 ) = 0 Formula used: L’Hopital Rule: suppose that we have one of the following case lim x → a f ( x ) g ( x ) = 0 0 or = ± ∞ ± ∞ Where a can be any real number, infinity or negative infinity. In these case we have, lim x → a f ( x ) g ( x ) = lim x → a f ' ( x ) g ' ( x ) And J n = ∫ 0 ∞ x n e − a x d x Calculation: ( a ) lim x → 0 f ( x ) = lim x → 0 x n e a x − 1 ( 0 0 f o r m ) Apply L ’Hopital rule lim x → 0 f ( x ) = lim x → 0 n x n − 1 a e x = 0 a = 0 = f ( 0 ) ⇒ f ( x ) is continuous at zero...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Chapter 8.7, Problem 94E

To determine

To show:

$(a)$ $f (x)$ is continuous at $x = 0$ where $f (x) = \frac{x^{n}}{e^{a x} - 1}$ for $x \neq 0, a > 0, n > 1.$

$(b)$ $\int_{0}^{\infty} f (x) d x$ converges.

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Chapter 8.7, Problem 93E

Chapter 8.7, Problem 95E

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Are the following statements true or false? If true give a proof, and if false give a counter-example: (a) Assume that that a sequence of continuous functions fn converges uniformly to a function f on the interval (a, b). This means that: ∀ε>0 ∃N>0 ∀n>N ∀x∈[a,b] : |fn(x)−f(x)|<ε. Then the function f is also continuous. (b)Assume that that a sequence of differentiable functions fn converges uniformly to a function f on the interval (a, b). Then the function f is also differentiable.

(c) A sequence (fn) of differentiable functions such that both (fn) and (fn )converge uniformly but f = lim fn is not differentiable at some point

In Example 2.4.1, show that fn(x) converges to f(x) pointwise on [0, 1]. (We haveseen the convergence at x = 0, 1.) For x ∈ [0, 1] and positive integer n, let fn(x) = xn.Then limn→∞ fn(x) = f(x),, wheref(x) = 0 if 0 ≤ x < 1,1 if x = 1.This example shows that the pointwise limit of a sequence of continuous functions need not be continuous.

Chapter 8 Solutions

CALCULUS (CLOTH)

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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( a ) f ( x ) is continuous at x = 0 where f ( x ) = x n e a x − 1 for x ≠ 0 , a > 0 , n > 1. ( b ) ∫ 0 ∞ f ( x ) d x converges.

Solution Summary: The author explains the L'Hopital Rule, where f (x) is continuous at x = 0 and d converges.

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To show:

$(a)$ $f (x)$ is continuous at $x = 0$ where $f (x) = \frac{x^{n}}{e^{a x} - 1}$ for $x \neq 0, a > 0, n > 1.$

$(b)$ $\int_{0}^{\infty} f (x) d x$ converges.

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Chapter 8 Solutions

( a ) f ( x ) is continuous at x = 0 where f ( x ) = x n e a x − 1 for x ≠ 0 , a &gt; 0 , n &gt; 1. ( b ) ∫ 0 ∞ f ( x ) d x converges.

Solution Summary: The author explains the L'Hopital Rule, where f (x) is continuous at x = 0 and d converges.

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To show: (a) f(x) is continuous at x=0 where f(x)=xneax−1 for x≠0,a>0,n>1. (b) ∫0∞f(x)dx converges.

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Chapter 8 Solutions

( a ) f ( x ) is continuous at x = 0 where f ( x ) = x n e a x − 1 for x ≠ 0 , a > 0 , n > 1. ( b ) ∫ 0 ∞ f ( x ) d x converges.

To show:

$(a)$ $f (x)$ is continuous at $x = 0$ where $f (x) = \frac{x^{n}}{e^{a x} - 1}$ for $x \neq 0, a > 0, n > 1.$

$(b)$ $\int_{0}^{\infty} f (x) d x$ converges.