Solution to Exercise 7.2-5© 1998 by Karl Hahn |
|
Step 1: The problem was to find the limit for
lim
h → 0
|
1 - cos(h) |
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
lim
h → 0
|
1 - cos2(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
lim
h → 0
|
sin2(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
lim
h → 0
|
1
|
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
lim
h → 0
|
1 1
|
So the answer is 1/2.
You can email me by clicking this button: