Karl's Calculus Tutor - Solution to Exercise 7.2-5

Solution to Exercise 7.2-5KCT logo

© 1998 by Karl Hahn

Step 1: The problem was to find the limit for 

          1 - cos(h)
    lim             
   h  > 0     h2
And the hint was to use the difference of squares to simplify it. That suggests that you multiply top and bottom by 1 + cos(h). Go ahead and do that, then click here.

Step 2: You should have gotten 

             1 - cos2(h)
    lim                    
   h  > 0  h2 (1 + cos(h) )
You should be able to see a trig identity that you can immediately apply to the numerator of this thing. Go ahead and apply it, and then click here.

Step 3: The trig identity that you should have applied is 1 - cos2(h) = sin2(h). When you make that substitution in the numerator, you get 

                sin2(h)
    lim                     
   h  > 0   h2 (1 + cos(h) )
Ok, you already know about the limit of sin(h)/h. You should be able to use that to infer what the limit of sin2(h)/h2 is. And in the limit, that leads to a cancellation. Go ahead and drop the terms that that eliminates. Then click here.

Step 4: The limit of sin2(h)/h2 is one as h goes to zero. So the sin2(h) term in the numerator cancels with the h2 term in the denominator. That leaves you with 

                1
    lim               
   h  > 0   1 + cos(h)
You know what cos(h) approaches as h approaches zero. So use that and take the limit. When you are done, click here.

As h gets closer and closer to zero, cos(h) gets closer and closer to one. So this limit looks more and more like 

              1       1
    lim            =   
   h  > 0   1 + 1     2
So the answer is 1/2.

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