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 27RE
To determine

Tofind:the limit of the function.

Expert Solution

Answer to Problem 27RE

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

  π=3.14159265.....

Explanation of Solution

Given:

  limx0tanπxln(1+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:

  limx0tanπxln(1+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.

  limx0tanπxln(1+x)

  =limx0ddx(tanπx)ddx(ln(1+x)) .

Finding derivative of numerator and denominator.

  =limx0πSec2(πx)1x+1 .

Taking limit of each term:

  πSec2(πlimx0x)limx01limx0x+limx01

Now by putting the value of x=0 in the above equation:

  πSec2(π.0)10+1 .

  π=3.14159265.....

Hence the limit of the function limx0tanπxln(1+x) is π=3.14159265.....

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!