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 4, Problem 32RE
To determine

Tofind:the limit of the function.

Expert Solution

Answer to Problem 32RE

Thelimit of the function limx0+x2lnx is 0 .

Explanation of Solution

Given:

  limx0+x2lnx .

Concept used:

If the function will be in the form of indeterminate (00) which is not valid.

In this kind of situation L Hospital’s Rule.

Which state that the limit of a quotient of the functions is equal to the limit of quotient of the derivative.

Calculation:

  limx0+x2lnx

The graph of the x and lnx

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4, Problem 32RE

  lnx goes to negative infinity while x approaches to 0 .

graph shows when approaches to 0

y also approaches to 0 .

  0.()=0.

limx0+x2lnx can be written as limx0+lnx1x2 .

By putting x=0 it will be in the form of indeterminate (00) .

In this kind of situation L Hospital’s Rule.

Which state that the limit of a quotient of the functions is equal to the limit of quotient of thederivative.

  limx0+lnx1x2

  =limx0ddx(lnx)ddx(1x2) .

Finding derivative of numeratordenominatorand taking limit of each term:

  limx01x2x3=limx0(x22)=0.

Hence the limit of the function limx0+x2lnx is 0 .

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