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 4.5, Problem 50E
To determine

To illustrate: L’Hospitals rule by ploting f(x)g(x) and f(x)g(x) near x=0 to check whether the limits are the same and calculate the exact value of the limit.

Expert Solution

Answer to Problem 50E

The exact value of limx02xsinxsecx1 is 4.

Explanation of Solution

Given:

The functions are f(x)=2xsinx and g(x)=secx1 .

Calculation:

The value of f(x)g(x)=2xsinxsecx1 .

The value of, f(x)g(x)=2xcosx+2sinxsecxtanx .

Use the online graphing calculator and draw the graphs of f(x)g(x)=2xsinxsecx1 and f(x)g(x)=2xcosx+2sinxsecxtanx on the same plane as shown below in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.5, Problem 50E

From Figure 1, it is identified that both curves approach to 4 as x approaches 0. Thus, f(x)g(x) and f(x)g(x) approaches the same limit.

Find the exact limit with the help of L’Hospital’s rule as follows.

Obtain the value of the function as x approaches 0 .

As x approaches 0, the numerator is,

2xsinx=20sin0=0

As x approaches 0, the denominator is,

secx1=sec01=11=0 .

Thus, limx02xsinxsecx1=00 is in an indeterminate form.

Therefore, apply L’Hospital’s Rule and obtain the limit.

limx02xsinxsecx1=limx02sinx+2xcosxsecxtanx

As x approaches 0, the numerator is,

2sinx+2xcosx=2sin0+20cos0=20+201=0+0=0

As x approaches 0, the denominator is,

secxtanx=sec0tan0=10=0

Thus, limx02sinx+2xcosxsecxtanx=00 is again in an indeterminate form.

Therefore, apply L’Hospital’s Rule again and obtain the limit.

limx02sinx+2xcosxsecxtanx=limx02cosx+2cosx2xsinxsec3x+secxtan2x=2cos0+2cos020sin0sec30+sec0tan20=2+201+10=4

Thus, limx02xsinxsecx1=4 .

Therefore, the exact value of limx02xsinxsecx1=4 .

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