# Tofind: the limit of the function. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### 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

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.....

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