# How do you find \lim _ { \theta \rightarrow 0} \frac { \cos ( 7\theta ) - 1} { \sin ( 4\theta ) }?

Dec 4, 2017

${\lim}_{\theta \to 0} \frac{\cos \left(7 \theta\right) - 1}{\sin} \left(4 \theta\right) = 0$

#### Explanation:

We first just try plugging in $0$ to see what we get:
$\frac{\cos \left(7 \cdot 0\right) - 1}{\sin} \left(4 \cdot 0\right) = \frac{1 - 1}{0} = \frac{0}{0}$

This results in the indeterminate form $\frac{0}{0}$, which doesn't help us.

There is however a trick you can use to evaluate limits of this indeterminate form. It's called L'Hôpital's Rule. It basically says that if both the top and the bottom tend to $0$, we can take the derivative of both the top and bottom, knowing that the limit will be the same. Formally, this is:
If ${\lim}_{x \to c} f \left(x\right) = 0$ and ${\lim}_{x \to c} g \left(x\right) = 0$ and ${\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)} = K$
Then ${\lim}_{x \to c} f \frac{x}{g} \left(x\right) = K$

Applying this to our case, we get:
lim_(theta->0)(cos(7theta)-1)/sin(4theta)=lim_(theta->0)(d/(d\theta)(cos(7theta)-1))/(d/(d\theta)(sin(4theta))

The derivatives can be computed using the chain rule:
$\frac{d}{d \setminus \theta} \left(\cos \left(7 \theta\right) - 1\right) = - 7 \sin \left(7 \theta\right)$

$\frac{d}{d \setminus \theta} \left(\sin \left(4 \theta\right)\right) = 4 \cos \left(4 \theta\right)$

Now we put back into the limit and evaluate at $0$:
${\lim}_{\theta \to 0} \frac{- 7 \sin \left(7 \theta\right)}{4 \cos \left(4 \theta\right)} = \frac{- 7 \sin \left(0\right)}{4 \cos \left(0\right)} = \frac{- 7 \cdot 0}{4 \cdot 1} = \frac{0}{4} = 0$