# Evaluate the limit: lim_(theta-> 0)(1-cos theta)/(sin 2theta)?

Jan 16, 2017

$\text{The Limit =} 0.$

#### Explanation:

We will use the Identity $: \sin 2 \theta = 2 \sin \theta \cos \theta$

Note that, $\frac{1 - \cos \theta}{\sin 2 \theta}$

$= \left\{\frac{1 - \cos \theta}{\sin 2 \theta}\right\} \left\{\frac{1 + \cos \theta}{1 + \cos \theta}\right\}$

$= \frac{1 - {\cos}^{2} \theta}{\left(\sin 2 \theta\right) \left(1 + \cos \theta\right)}$

$= {\sin}^{2} \frac{\theta}{\left(2 \sin \theta \cos \theta\right) \left(1 + \cos \theta\right)}$

$= \frac{\sin \theta}{2 \cos \theta \left(1 + \cos \theta\right)}$

$= \frac{\sin 0}{\left(2\right) \left(\cos 0\right) \left(1 + \cos 0\right)} = \frac{0}{4}$

$\therefore \text{ the Reqd. Lim.=} 0$

Jan 16, 2017

The limit is 0.

#### Explanation:

There are (at least) two ways to evaluate this limit. The first way is to apply l'Hopital's Rule, which says

If both $f \left(x\right) \to 0 \text{ and " g(x) -> 0 " as } x \to 0$

then ${\lim}_{x \to 0} \frac{f \left(x\right)}{g \left(x\right)} \text{ "=" } {\lim}_{x \to 0} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

Since both $1 - \cos \theta$ and $\sin 2 \theta$ approach 0 as $\theta \to 0$, we can apply l'Hopital's rule to get:

$\textcolor{w h i t e}{=} {\lim}_{\theta \to 0} \frac{1 - \cos \theta}{\sin 2 \theta}$

$= {\lim}_{\theta \to 0} \sin \frac{\theta}{2 \cos \left(2 \theta\right)}$

After this step, we no longer have a "division by zero" limit, so we can just plug in 0 for $\theta$ to get our answer:

$= \sin \frac{0}{2 \cos \left[2 \left(0\right)\right]}$

$= \frac{0}{2 \left(1\right)}$

$= 0$

Another way possible is to use $\sin 2 \theta = 2 \sin \theta \cos \theta$, and also multiply both numerator and denominator by $1 + \cos \theta$ to achieve

$\textcolor{w h i t e}{=} {\lim}_{\theta \to 0} \frac{1 - \cos \theta}{\sin 2 \theta}$

$= {\lim}_{\theta \to 0} \frac{1 - \cos \theta}{2 \sin \theta \cos \theta} \cdot \frac{1 + \cos \theta}{1 + \cos \theta}$

$= {\lim}_{\theta \to 0} \frac{1 - {\cos}^{2} \theta}{2 \sin \theta \cos \theta \left(1 + \cos \theta\right)}$

By the Pythagorean identities, we have

$= {\lim}_{\theta \to 0} \frac{{\sin}^{2} \theta}{2 \sin \theta \cos \theta \left(1 + \cos \theta\right)}$

$= {\lim}_{\theta \to 0} \frac{\sin \theta}{2 \cos \theta \left(1 + \cos \theta\right)}$

Once again, we no longer have a "division by zero", so we can directly substitute 0 for $\theta$ to get

$= \frac{\sin \left(0\right)}{2 \cos \left(0\right) \left[1 + \cos \left(0\right)\right]}$

$= \frac{0}{2 \left(1\right) \left[1 + 1\right]}$

$= 0$.

Here is a graph of the function $f \left(\theta\right) = \frac{1 - \cos \theta}{\sin 2 \theta}$:
graph{(1-cos x)/(sin(2x)) [-10, 10, -5, 5]}