# How do you find the limit of tan(4alpha)/sin(2alpha) as alpha->0?

Mar 2, 2017

$2$

#### Explanation:

${\lim}_{\alpha \to 0} \frac{\tan 4 \alpha}{\sin 2 \alpha}$

This is in $\frac{0}{0}$ indeterminate form so L'Hopital is a possibility.

We do it algebraically, using:

$\sin 2 z = 2 \sin z \cos z$, and

$\cos 2 z = {\cos}^{2} z - {\sin}^{2} z$ and variations of this and the Pythagorean ID.

We can say:

$= {\lim}_{\alpha \to 0} \frac{\sin 4 \alpha}{\cos 4 \alpha} \frac{1}{\sin 2 \alpha}$

$= {\lim}_{\alpha \to 0} \frac{2 \sin 2 \alpha \cos 2 \alpha}{2 {\cos}^{2} 2 \alpha - 1} \frac{1}{\sin 2 \alpha}$

$= {\lim}_{\alpha \to 0} \frac{2 \cos 2 \alpha}{2 {\cos}^{2} 2 \alpha - 1}$

These are both continuous at $\alpha = 0$ so we can plug in $\alpha = 0$ to get:

$= {\lim}_{\alpha \to 0} \frac{2 \cdot 1}{2 \cdot 1 - 1} = 2$

Mar 2, 2017

To solve without using derivatives (l'Hospital's Rule) see below.

#### Explanation:

$\tan \left(4 \alpha\right) = \sin \frac{4 \alpha}{\cos} \left(4 \alpha\right)$

Use the sine of a double angle formula to rewrite this

$= \frac{2 \sin \left(2 \alpha\right) \cos \left(2 \alpha\right)}{\cos} \left(4 \alpha\right)$

So

$\tan \frac{4 \alpha}{\sin} \left(2 \alpha\right) = \frac{2 \sin \left(2 \alpha\right) \cos \left(2 \alpha\right)}{\sin \left(2 \alpha\right) \cos \left(4 \alpha\right)}$

$= \frac{2 \cos \left(2 \alpha\right)}{\cos \left(4 \alpha\right)}$

Evaluating the limit as $\alpha \rightarrow 0$ we get

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

Mar 2, 2017

${\lim}_{\alpha \rightarrow 0} \tan \frac{4 \alpha}{\sin} \left(2 \alpha\right) = 2$

#### Explanation:

We can also use the standard limit ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$. This can be inverted to say that ${\lim}_{x \rightarrow 0} \frac{x}{\sin} \left(x\right) = 1$.

We can rearrange the given limit to make these appear:

${\lim}_{\alpha \rightarrow 0} \tan \frac{4 \alpha}{\sin} \left(2 \alpha\right) = {\lim}_{\alpha \rightarrow 0} \sin \left(4 \alpha\right) \frac{1}{\cos} \left(4 \alpha\right) \frac{1}{\sin} \left(2 \alpha\right)$

Multiplying by $\frac{\alpha}{\alpha}$:

$= {\lim}_{\alpha \rightarrow 0} \sin \frac{4 \alpha}{\alpha} \frac{1}{\cos} \left(4 \alpha\right) \frac{\alpha}{\sin} \left(2 \alpha\right)$

Multiplying by $\frac{2 \left(2\right)}{4}$:

$= 2 \left({\lim}_{\alpha \rightarrow 0} \sin \frac{4 \alpha}{4 \alpha} \frac{1}{\cos} \left(4 \alpha\right) \frac{2 \alpha}{\sin} \left(2 \alpha\right)\right)$

We can split up the limit. We also see that ${\lim}_{\alpha \rightarrow 0} \sin \frac{4 \alpha}{4 \alpha}$ and ${\lim}_{\alpha \rightarrow 0} \frac{2 \alpha}{\sin} \left(2 \alpha\right)$ are in the form ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$.

$= 2 \cdot {\lim}_{\alpha \rightarrow 0} \frac{2 \alpha}{\sin} \left(2 \alpha\right) \cdot {\lim}_{\alpha \rightarrow 0} \frac{1}{\cos} \left(4 \alpha\right) \cdot {\lim}_{\alpha \rightarrow 0} \frac{2 \alpha}{\sin} \left(2 \alpha\right)$

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

$= 2$