# 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 34RE
To determine

## Tofind:the limit of the function.

Expert Solution

Thelimit of the function limxπ2(Tanx)Cosx is 1 .

### Explanation of Solution

Given:

limxπ2(Tanx)Cosx .

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:

limxπ2(Tanx)Cosx .

Let l is the limit which is equal to the limxπ2(Tanx)Cosx .

l=limxπ2(Tanx)Cosx .

By taking log to both side of the function.

logl=limxπ2-logtanxsecx .

By putting direct x=π2 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.

logl=limxπ2-logtanxsecx

logl=limxπ2-ddx(logtanx)ddx(secx)

logl=limxπ2-1tanxsec2xsecxtanx .

logl=limxπ2-secxtan2x .

By putting again x=π2 the function will be in the form of indeterminate (00) which is not valid.

In this kind of situation L Hospital’s Rule.

logl=limxπ2-secxtan2x .

Finding derivative of numerator and denominator

logl=limxπ2-secxtanx2tanxsec2x=12cosx=0.

logl=0.

l=1 .

Hence the limit of the function limxπ2(Tanx)Cosx is 1 .

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