# derive-it.com

## A Limit Involving the Cosine Function

Now that several limit properties have been proven, it is possible for me to evaluate

$\lim_{\alpha \rightarrow 0} \frac{1 – \cos \alpha}{\alpha}$.

To do this, I follow the steps in Reference . However, I am going to constrain $\alpha$, in radians, to be greater than or equal to zero, so that I do not need to deal with the issue of negative angles and how they are used in trigonometric identities proven with right triangles and nonnegative angles. I know that the limit definition requires negative and positive numbers if $x \rightarrow 0$, but I am going to ignore the negative numbers, at least for now. This deviation from the traditional limit definition is only a slight modification; in fact, there might be a definition of a “right-handed” limit. To denote this constraint, I will write $\lim_{\alpha \rightarrow 0^+}$ instead of $\lim_{\alpha \rightarrow 0}$. So, I am actually evaluating $\lim_{\alpha \rightarrow 0^+} \frac{1 – \cos \alpha}{\alpha}$.

$2\sin^2 \frac{\alpha}{2} = 1-\cos \alpha$.

Using the previous equation, the numerator of the limit can be replaced:

$\lim_{\alpha \rightarrow 0^+} \frac{1 – \cos \alpha}{\alpha} = \lim_{\alpha \rightarrow 0^+} \frac{2\sin^2 \frac{\alpha}{2}}{\alpha}$.

Rewriting,

$= \lim_{\alpha \rightarrow 0^+} \sin \frac{\alpha}{2} \frac{\sin \frac{\alpha}{2}}{\frac{\alpha}{2} }$.

Using this property,

$= \lim_{\alpha \rightarrow 0^+} \sin \frac{\alpha}{2} \lim_{\alpha \rightarrow 0^+} \frac{\sin \frac{\alpha}{2}}{\frac{\alpha}{2} }$.

By substituting $0$ for $\alpha$ in $\sin \frac{\alpha}{2}$ and by noticing that $\sin \frac{\alpha}{2}$ decreases as $\frac{\alpha}{2}$ decreases, it is clear that $\lim_{\alpha \rightarrow 0^+} \sin \frac{\alpha}{2} = 0$.

The other limit, $\lim_{\alpha \rightarrow 0^+} \frac{\sin \frac{\alpha}{2}}{\frac{\alpha}{2} }$, is evaluated by using the reasoning in this post . In particular, the limit was considered by noticing:

As the angle decreases, does the arc length’s value become closer to the $y$-value? Geometrically, the answer is yes, and indeed at $\theta=0$ radians the $y$-value and the arc length are the same value of zero. From this information I conclude that $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$.

This conclusion was made before I realized there was an issue with negative angles, but the logic still applies. Here is a diagram. As $\alpha$ decreases as in the limit under consideration, the value of the arc length corresponding to $\frac{\alpha}{2}$ radians becomes closer to the $y$-value corresponding to $\frac{\alpha}{2}$ radians. To conclude this, it is helpful to analyze how the shape of the arc length changes as the angle changes.

Indeed, at $\alpha = 0$ radians, the $y$-value corresponding to $\frac{\alpha}{2}= 0$ radians and the arc length corresponding to $\frac{\alpha}{2}= 0$ radians have the same value of zero. From this information, I conclude that

$\lim_{\alpha \rightarrow 0^+} \frac{\sin \frac{\alpha}{2}}{\frac{\alpha}{2} } = 1$.

Therefore,

$\lim_{\alpha \rightarrow 0^+} \sin \frac{\alpha}{2} \lim_{\alpha \rightarrow 0^+} \frac{\sin \frac{\alpha}{2}}{\frac{\alpha}{2} } = 0 * 1 = 0$.

so

$\lim_{\alpha \rightarrow 0^+} \frac{1 – \cos \alpha}{\alpha} = 0$.

References:

 Morris Kline. Calculus, An Intuitive and Physical Approach. Dover Publications, 1998.