2) For lim??→0+(5??)4??, do not evaluate!a) Which indeterminant form is the expression?b) Set up the limit in order that L’Hopital’s Rule can be applied, state the new indeterminant form. 2) For lim (5x)4*, do not evaluate!x→0+a) Which indeterminant form is the expression?b) Set up the limit in order that L’Hopital’s Rule can be applied, state the new indeterminant form.

a) The given limit expression =limx→0+5x4x=00 form

Answered: 2) For lim (5x)4*, do not evaluate!…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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2) For lim??→0+(5??)4??, do not evaluate!a) Which indeterminant form is the expression?b) Set up the limit in order that L’Hopital’s Rule can be applied, state the new indeterminant form.

2) For lim (5x)4*, do not evaluate!
x→0+
a) Which indeterminant form is the expression?
b) Set up the limit in order that L’Hopital’s Rule can be applied, state the new indeterminant form.

Expert Solution

Step 1

The given limit expression

$\underset{x \to 0^{+}}{= \lim} {(5 x)}^{4 x} = {(0)}^{0} form$

Step 2

The above limit can be written in exponential form as

$e^{\lim_{x \to 0} \ln {(5 x)}^{(4 x)}} = e^{\lim_{x \to 0} (4 x) \ln (5 x)} = e^{4 \lim_{x \to 0} \frac{\ln (5 x)}{\frac{1}{x}}}$

So the limit $\frac{\ln (5 x)}{\frac{1}{x}} = \frac{\infty}{\infty} form$

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