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 28RE
To determine

Tofind:the limit of the function.

Expert Solution

Answer to Problem 28RE

The limit of the function limx0tanπxln(1+x) is 0 .

Explanation of Solution

Given:

  limx01-Cosxx2+x .

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:

  limx01-Cosxx2+x .

By putting direct x=0 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 there derivative.

  limx01-Cosxx2+x

  =limx0ddx(1-Cosx)ddx(x2+x) .

Finding derivative of numerator and denominator.

  =limx0Sinx2x+1 .

  =limx0Sinxx(2+1x)=limx01.12+=0.

Hence the limit of the function limx01-Cosxx2+x is 0 .

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