# How do you find the limit of  (sin (4x)) / (tan(5x))  as x approaches 0?

Mar 16, 2016

Use Algebra, trigonometry and the fundamental trigonometric limit.

#### Explanation:

${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$

We can use this to find ${\lim}_{x \rightarrow 0} \sin \frac{7 x}{x}$ because

${\lim}_{x \rightarrow 0} \sin \frac{7 x}{x} = {\lim}_{x \rightarrow 0} \frac{7}{1} \sin \frac{7 x}{7 x}$

$= 7 {\lim}_{x \rightarrow 0} \sin \frac{7 x}{7 x}$

Now, with $7 x = \theta$ we see that ${\lim}_{x \rightarrow 0} \sin \frac{7 x}{7 x} = 1$

So we finish with

$7 {\lim}_{x \rightarrow 0} \sin \frac{7 x}{7 x} = 7 \left(1\right) = 7$#

We'll also need ${\lim}_{\theta \rightarrow 0} \cos \theta = 1$ (Cosine is continuous at $0$.)

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

Here is the solution

$\sin \frac{4 x}{\tan} \left(5 x\right) = \sin \frac{4 x}{\sin \frac{5 x}{\cos} \left(5 x\right)}$

$= \sin \frac{4 x}{1} \cos \left(5 x\right) \frac{1}{\sin} \left(5 x\right)$

$= \sin \frac{4 x}{4} x \cdot \frac{4}{1} \cos \left(5 x\right) 5 \frac{x}{\sin} \left(5 x\right) \cdot \frac{1}{5}$

$= \frac{4}{5} \left[\sin \frac{4 x}{4} x\right] \left[\cos \left(5 x\right)\right] \left[5 \frac{x}{\sin} \left(5 x\right)\right]$

Taking limits as $x \rightarrow 0$ we get,

$= \frac{4}{5} \left[1\right] \left[1\right] \left[1\right] = \frac{4}{5}$