# How do you find the limit of (sin(7 x)) / (tan(4 x)) as x approaches 0?

Mar 3, 2016

7/4

#### Explanation:

Let $f \left(x\right) = \sin \frac{7 x}{\tan} \left(4 x\right)$

$\implies f \left(x\right) = \sin \frac{7 x}{\sin \frac{4 x}{\cos} \left(4 x\right)}$

$\implies f \left(x\right) = \sin \frac{7 x}{\sin} \left(4 x\right) \cdot \cos \left(4 x\right)$

$\implies f ' \left(x\right) = {\lim}_{x \to 0} \left\{\sin \frac{7 x}{\sin} \left(4 x\right) \cdot \cos \left(4 x\right)\right\}$

$\implies f ' \left(x\right) = {\lim}_{x \to 0} \left\{\frac{7 \cdot \sin \frac{7 x}{7 x}}{4 \cdot \sin \frac{4 x}{4 x}} \cdot \cos \left(4 x\right)\right\}$

$\implies f ' \left(x\right) = \frac{7}{4} {\lim}_{x \to 0} \left\{\frac{\sin \frac{7 x}{7 x}}{\sin \frac{4 x}{4 x}} \cdot \cos \left(4 x\right)\right\} = \frac{7}{4} \frac{{\lim}_{x \to 0} \sin \frac{7 x}{7 x}}{{\lim}_{x \to 0} \sin \frac{4 x}{4 x}} \cdot {\lim}_{x \to 0} \cos \left(4 x\right) = \frac{7}{4} \cdot \frac{1}{1} \cdot \cos \left(4 \cdot 0\right) = \frac{7}{4} \cdot \cos 0 = \frac{7}{4} \cdot 1 = \frac{7}{4}$