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!…

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.

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter3: Functions

Section3.3: Rates Of Change And Behavior Of Graphs

Problem 2SE: If a functionfis increasing on (a,b) and decreasing on (b,c) , then what can be said about the local...

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Concept explainers

Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

Question

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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