Expert: Ahmed Salami - 4/30/2009

Question
Hi, So I have a worksheet that is just to help us get started before we start our next lesson and I was just really confused on it because I can't get one part of the sheet without getting something else. So anything helps!

1. Find lim   cos(theta)-1
  (theta)->0 ------------
               (theta)
using the following outline:

a) multiply by cos(theta)+1
              ------------
              cos(theta)+1


b) Use the Pythagorean identity to simplify the numerator


c) rewrite the fraction as a product of two fractions, where one of the fractions is sin(theta)
                             ----------
                              (theta)


d) use limits laws to find  lim      cos(theta)-1
                         (theta)->0 ------------
                                       (theta)  

Answer
Hi Ray,
I'll be representing theta by # and writing the limit as theta approaches zero as just 'lim'.
a)(cos# - 1)/# = (cos# - 1)/# . (cos# + 1)/(cos# + 1)
              = (cos# - 1)(cos# + 1)/#(cos# + 1)
              = [(cos#)^2 + cos# - cos# - 1] / #(cos# + 1)
              = [(cos#)^2 - 1] / #(cos# + 1)
b)Using the Pythagorean identity,
(cos#)^2 + (sin#)^2 = 1
(cos#)^2 = 1 - (sin#)^2
And so the numerator becomes
(cos#)^2 - 1 = 1 - (sin#)^2 - 1
            = -(sin#)^2
c)Rewriting, we have
(cos# - 1)/# = -(sin#)^2 / #(cos# + 1)
            = [(sin#) / #].[-sin# / (cos# + 1)]
d)By the laws of limits,
lim AB = limA . limB
Therefore,
lim (cos# - 1)/# = lim [(sin#) / #].[-sin# / (cos# + 1)]
                = lim [(sin#) / #].lim [-sin# / (cos# + 1)]
but, as # approaches zero
lim [(sin#) / #] = 1   (This is a standard solution gotten by the Squeeze Theorem)
lim [-sin# / (cos# + 1)] = [-sin0 / (cos0 + 1)]
                        = 0/1+1
                        = 0/2
                        = 0
Finally,
lim (cos# - 1)/# = 1 . 0
                = 0

Hope it helps you.

Regards