Expert: Paul Klarreich - 9/16/2007

Question
Hi, Im learning the squeeze/sandwich theorem and am having trouble with proving a particular limit.

Using the squeeze theorem I need to prove that lim x-> 0 (1-cosx)/x = 0.

My instructor worked out the theorem using lim x-> 0 (sinx)/x = 1 but Im having trouble relating the two.

It would be awesome if you could help me.

-Micah

Answer
Questioner:   Micah
Category:  Calculus
Private:  No
 
Subject:  Calculus, Squeeze Theorem
Question:  Hi, Im learning the squeeze/sandwich theorem and am having trouble with proving a particular limit.

Using the squeeze theorem I need to prove that lim x-> 0 (1-cosx)/x = 0.

My instructor worked out the theorem using lim x-> 0 (sinx)/x = 1 but Im having trouble relating the two.

It would be awesome if you could help me.

-Micah
........................................
Hi, Micah,

You have to prove this limit:

     1 - cos x
lim   --------- = 0
x->0      x

I'm not sure what you were asking, exactly.  I'm sure your instructor provided a proof along these lines:

     1 - cos x
lim   --------- =
x->0      x

     1 - cos x  1 + cos x
lim   ---------  ---------- =  << rationalizing
x->0      x      1 + cos x

     1 - cos^2(x)
lim   -------------   
x->0  x(1 + cos x)

       sin^2(x)       << use a trig identity here
lim   -------------   
x->0  x(1 + cos x)

     sin x  sin x       << separate the factors
lim   -------------   
x->0  x(1 + cos x)

     sin x    sin x       << separate the fractions
lim   ----- -----------   
x->0    x   (1 + cos x)

     sin x          sin x       
lim   -----  lim   -----------  << lim of prod = prod of limits
x->0    x    x->0 (1 + cos x)

                   sin x       
       1  lim   -----------  << known limit
          x->0  (1 + cos x)

                   sin 0       
       1  lim   -----------
          x->0  (1 + cos 0)

                   0       
       1  lim   -------
          x->0  (1 + 1)

       1 * 0

Now if you have to prove this identity from scratch, that's another matter.  If you must, let me know and I'll try it.