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

Sep 24, 2015

The usual procedure is to use the squeeze theorem (and some geometry/trigonometry) to prove that ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$

Then use that result together with $\frac{1 - \cos x}{x} = {\sin}^{2} \frac{x}{x} \left(1 + \cos x\right) = \sin \frac{x}{x} \sin \frac{x}{1 + \cos x}$ along with continuity of sine and cosine at $0$ to get
${\lim}_{x \rightarrow 0} \sin \frac{x}{x} \sin \frac{x}{1 + \cos x} = 1 \cdot \frac{0}{2} = 0$.

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

$\cos x < \sin \frac{x}{x} < 1$

And for small positive $x$, we have $\sin \frac{x}{1 + \cos x} > 0$, so we can multiply without changing the inequalities:

$\cos x \sin \frac{x}{1 + \cos x} < \sin \frac{x}{x} \sin \frac{x}{1 + \cos x} < \sin \frac{x}{1 + \cos x}$

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

$\cos x \sin \frac{x}{1 + \cos x} < \frac{1 - \cos x}{x} < \sin \frac{x}{1 + \cos x}$

Observe that

${\lim}_{x \rightarrow {0}^{+}} \left(\cos x \sin \frac{x}{1 + \cos x}\right) = 1 \cdot \frac{0}{2} = 0$

and

${\lim}_{x \rightarrow {0}^{+}} \sin \frac{x}{1 + \cos x} = \frac{0}{2} = 0$

So, by the squeeze theorem,

${\lim}_{x \rightarrow {0}^{+}} \frac{1 - \cos x}{x} = 0$

For small negative $x$, we keep the inequality: $\cos x < \sin \frac{x}{x} < 1$, but for these $x$, we have

$\sin \frac{x}{1 + \cos x} < 0$, so when we multiply we do need to change the inequalities to:

$\cos x \sin \frac{x}{1 + \cos x} > \frac{1 - \cos x}{x} > \sin \frac{x}{1 + \cos x}$

We can still use the squeeze theorem to get:

${\lim}_{x \rightarrow {0}^{-}} \frac{1 - \cos x}{x} = 0$

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.)