How do you use the Squeeze Theorem to find #lim (1-cos(x))/x# as x approaches zero?

1 Answer
Sep 24, 2015

The usual procedure is to use the squeeze theorem (and some geometry/trigonometry) to prove that #lim_(xrarr0)sinx/x=1#

Then use that result together with #(1-cosx)/x = sin^2x/x(1+cosx) = sinx/x sinx/(1+cosx)# along with continuity of sine and cosine at #0# to get
#lim_(xrarr0)sinx/x sinx/(1+cosx) = 1 * 0/2 =0#.

So we can use the same geometric arguments to get the same bounds on sinx/x for small positive #x#:

#cosx < sinx/x < 1#

And for small positive #x#, we have #sinx/(1+cosx) > 0#, so we can multiply without changing the inequalities:

#cosx sinx/(1+cosx) < sinx/x sinx/(1+cosx) < sinx/(1+cosx) #

Using the trigonomtry referred to above, we can rewrite the midle expression to get

#cosx sinx/(1+cosx) < (1-cosx)/x < sinx/(1+cosx) #

Observe that

#lim_(xrarr0^+)(cosx sinx/(1+cosx)) = 1*0/2 = 0#

and

#lim_(xrarr0^+)sinx/(1+cosx) = 0/2=0#

So, by the squeeze theorem,

#lim_(xrarr0^+)(1-cosx)/x = 0#

For small negative #x#, we keep the inequality: #cosx < sinx/x < 1#, but for these #x#, we have

#sinx/(1+cosx) < 0#, so when we multiply we do need to change the inequalities to:

#cosx sinx/(1+cosx) > (1-cosx)/x > sinx/(1+cosx) #

We can still use the squeeze theorem to get:

#lim_(xrarr0^-)(1-cosx)/x = 0#

Because the left and right limits are both #0#, the limit is #0#.

(This feels very artificial to me. Perhaps because I am more familiar with the common approach mentioned at the beginning of this answer. or perhaps because it is artificial. I don't know.)