Karl's Calculus Tutor - Solution to Exercise 8.1-3

Solution to Exercise 8.1-3KCT logo

© 1998 by Karl Hahn

The problem was to find the limit of 

           arcsin(x) - p/2
    lim                   
   x  > 1      Ö1 - x2
Step 1: Determine if both numerator and denominator go to zero. We know that arcsin(1) = p/2, so clearly the numerator goes to zero as x approaches 1. In the denominator, it's clear that x2 goes to 1 as x approaches 1. So the difference inside the square root function must go zero. Square root is a continuous function for nonnegative x, and the square root of zero is zero. So as x approaches 1, the denominator must go to zero as well. So this limit qualifies for L'Hopital's rule.

Step 2: Take the derivatives of the numerator and denominator. We discussed in section 7.2 how the derivative of arcsin(x) was 

      1
          
   Ö1 - x2
Since the p/2 is a constant, its derivative is zero. So the derivative of the numerator is also 
      1
          
   Ö1 - x2
To take the derivative of the denominator, 
    ______
   Ö1 - x2
you will need to use the chain rule. When you do you will get 
   1   -2x          -x
              =         
   2 Ö1 - x2     Ö1 - x2
Make sure you can take this derivative on your own.

Step 3: Apply L'Hopital's Rule. That means 

                                                1
                                                    
            arcsin(x) - p/2                  Ö1 - x2
    lim                      =   lim                    
   x  > 1      Ö1 - x2          x  > 1         -x
                                                    
                                             Ö1 - x2
The 
    ______
   Ö1 - x2)
terms in the numerator and denominator cancel, and you are left with 
            arcsin(x) - p/2              -1
    lim                      =   lim        
   x  > 1      Ö1 - x2          x  > 1    x
When you take the limit of -1/x as x approaches 1 you get a limit of -1, which is the answer. You can experiment on your calculator with taking the original function of x's that are very close to 1 to confirm this limit.

Return to Main Text

email me at hahn@netsrq.com