How do you find the limit of  (sin 3x)/ (sin 4x) as x approaches 0?

Apr 24, 2016

Use ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$ and some algebra.

Explanation:

Note: because ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$, we also have ${\lim}_{\theta \rightarrow 0} \frac{\theta}{\sin} \theta = 1$

Rewrite the expression to use the limits noted above.

$\frac{\sin 3 x}{\sin 4 x} = \frac{\sin 3 x}{1} \cdot \frac{1}{\sin 4 x}$

$= \left[\frac{3 x}{1} \frac{\sin 3 x}{3 x}\right] \cdot \left[\frac{1}{4 x} \frac{4 x}{\sin 4 x}\right]$

$= \frac{3 x}{4 x} \sin \frac{3 x}{3 x} \frac{4 x}{\sin} \left(4 x\right)$

$= \frac{3}{4} \left[\frac{\sin 3 x}{3 x} \frac{4 x}{\sin 4 x}\right]$

Now, as $x \rightarrow 0$, $\left(3 x\right) \rightarrow 0$ so $\frac{\sin 3 x}{3 x} \rightarrow 1$. (Using $\theta = 3 x$)

And, as $x \rightarrow 0$, $\left(4 x\right) \rightarrow 0$ so $\frac{4 x}{\sin 4 x} \rightarrow 1$. (Using $\theta = 4 x$)

Therefore the limit is $\frac{3}{4}$.

${\lim}_{x \rightarrow 0} \frac{\sin 3 x}{\sin 4 x} = {\lim}_{x \rightarrow 0} \frac{3}{4} \frac{\sin 3 x}{3 x} \frac{4 x}{\sin 4 x}$

$= \frac{3}{4} {\lim}_{x \rightarrow 0} \frac{\sin 3 x}{3 x} {\lim}_{x \rightarrow 0} \frac{4 x}{\sin 4 x}$

$= \frac{3}{4} \left(1\right) \left(1\right) = \frac{3}{4}$