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

Question

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

Isabella Ackerman · Accepted Answer

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) #

With the help of the previously mentioned trigonomtry, we can rewrite the midle expression to obtain

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

Note that

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

and

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

In light of 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) #

The squeeze theorem can still be applied to obtain:

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

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

(Perhaps because I am more accustomed to the standard procedure that was stated at the opening of this response, or maybe just because it is artificial—I'm not sure.) This seems really fake to me.

Christopher Allers · Answer

undefined To use the Squeeze Theorem to find the limit of (1-cos(x))/x as x approaches zero, we can start by observing that -1 ≤ cos(x) ≤ 1 for all x.

Since 1 - cos(x) ≥ 0, we can divide both sides of the inequality by x (assuming x ≠ 0) to get:

(1 - cos(x))/x ≥ 0/x

Simplifying this expression, we have:

(1 - cos(x))/x ≥ 0

Now, let's consider the upper bound. We know that cos(x) ≤ 1, so we can divide both sides of the inequality by x (assuming x ≠ 0) to get:

(1 - cos(x))/x ≤ 1/x

Simplifying this expression, we have:

(1 - cos(x))/x ≤ 1/x

Now, as x approaches zero, both the lower bound and upper bound approach zero. Therefore, by the Squeeze Theorem, the limit of (1 - cos(x))/x as x approaches zero is also zero.