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

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.)