Show that $ \lim_{x \to 0} \frac{\sin\, x}{x} = 1$

To motivate the limit computation of this section, using Sage we compute a number of values of the function $ \frac{\sin\, x}{x}$, as $ x$ gets closer and closer to 0:

$ x$ 0.5000 0.2500 0.1250 0.06250 0.03125
$ \frac{\sin(x)}{x}$ 0.9589 0.9896 0.9974 0.9994 0.9998

Indeed, if we refer to the table in §1.4, it will be seen that for all angles less than $ 10^o$ the angle in radians and the sine of that angle are equal to three decimal places. To compute the table of values above using Sage, simply use the following commands.

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: f = lambda x: sin(x)/x
sage: R = RealField(15)
sage: L = [1/2^i for i in range(1,6)]; L
[1/2, 1/4, 1/8, 1/16, 1/32]
sage: [R(x) for x in L]
[0.5000, 0.2500, 0.1250, 0.06250, 0.03125]
sage: [R(f(x)) for x in L]
[0.9589, 0.9896, 0.9974, 0.9994, 0.9998]

From this we may well suspect that $ \lim_{x \to 0} \frac{\sin\, x}{x} = 1$.

Let $ O$ be the center of a circle whose radius is unity.

Let $ \operatorname{arc}\ AM = \operatorname{arc}\ AM' = x$, and let $ MT$ and $ M'T$ be tangents drawn to the circle at $ M$ and $ M'$. From Geometry (see Figure 3.12),

Figure 3.12: Comparing $ x$ and $ \sin(x)$ on the unit circle.
\includegraphics[height=5cm,width=6cm]{circle-tangent.eps}

we have

$\displaystyle MPM' < MAM' < MTM';
$

or $ 2 \sin\, x < 2x < 2 \tan\ x$. Dividing through by $ 2 \sin\, x$, we get

$\displaystyle 1 < \frac{x}{\sin\, x} < \frac{1}{\cos\ x}.
$

If now $ x$ approaches the limit zero,

$\displaystyle \lim_{x \to 0} \frac{x}{\sin\, x}
$

must lie between the constant $ 1$ and $ \lim_{x \to 0} \frac{1}{\cos\ x}$, which is also $ 1$. Therefore $ \lim_{x \to 0} \frac{x}{\sin\, x} = 1$, or, $ \lim_{x \to 0} \frac{\sin\, x}{x} = 1$ Theorem 3.8.3. $ \qedsymbol$

It is interesting to note the behavior of this function from its graph, the locus of equation

$\displaystyle y = \frac{\sin\, x}{x}
$

Figure: The function $ \frac{\sin(x)}{x}$.
\includegraphics[height=4cm,width=8cm]{limit_proof2.eps}

Although the function is not defined for $ x = 0$, yet it is not discontinuous when $ x = 0$ if we define $ \frac{\sin\, 0}{0} = 1$ (see Case II in §3.6).

Finally, we show how to use the Sage command limit to compute the limit above3.5.

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: limit(sin(x)/x,x=0)
1

david joyner 2008-11-22