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

Oct 19, 2015

For a non-rigorous proof, please see below.

#### Explanation:

For a positive central angle of $x$ radians ($0 < x < \frac{\pi}{2}$) (not degrees)

Source:

The geometric idea is that

$\text{Area of "Delta KOA < "Area of " "Sector KOA" < "Area of } \Delta L O A$

$\text{Area of } \Delta K O A = \frac{1}{2} \left(1\right) \left(\sin x\right) \setminus \setminus \setminus$ ($\frac{1}{2} \text{base"*"height}$)

$\text{Area of " "Sector KOA} = \frac{1}{2} {\left(1\right)}^{2} x \setminus \setminus \setminus$ ($x$ is in radians)

$\text{Area of } \Delta L O A = \frac{1}{2} \tan x \setminus \setminus \setminus$ ($A L = \tan x$)

So we have:

$\sin \frac{x}{2} < \frac{x}{2} < \tan \frac{x}{2}$

For small positive $x$, we have $\in x > 0$ so we can multiply through by $\frac{2}{\sin} x$, to get

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

So

$\cos x < \sin \frac{x}{x} < 1$ for $0 < x < \frac{\pi}{2}$.

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

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

We also have, for these small $x$, $\sin \left(- x\right) = - \sin x$, so $\frac{- x}{\sin} \left(- x\right) = \frac{x}{\sin} x$ and $\cos \left(- x\right) = \cos x$, so

$\cos x < \sin \frac{x}{x} < 1$ for $- \frac{\pi}{2} < x < 0$.

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

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

Since both one sided limits are $1$, the limit is $1$.

Note

This proof uses the fact that ${\lim}_{x \rightarrow 0} \cos x = 1$. That can also be stated "the cosine function is continuous at $0$".

That fact can be proved from the fact that ${\lim}_{x \rightarrow 0} \sin x = 0$. (The sine function is continuous at $0$.)
Which can be proved using the squeeze theorem in a argument rather like the one used above.

Furthermore: Using both of those facts we can show that the sine and cosine functions are continuous at every real number.