# How do you evaluate the limit sin(3x)/x as x approaches 0?

Oct 22, 2016

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

#### Explanation:

One way to use ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$ is to use $\theta = 3 x$

But now we need $3 x$ in the denominator.

No problem, multiply by $\frac{3}{3}$

${\lim}_{x \rightarrow 0} \sin \frac{3 x}{x} = {\lim}_{x \rightarrow 0} 3 \cdot \sin \frac{3 x}{\left(3 x\right)}$

As $x \rightarrow 0$, s also $3 x \rightarrow 0$. We can substitute to get

${\lim}_{\theta \rightarrow 0} 3 \cdot \sin \frac{\theta}{\theta} = 3 \cdot 1 = 3$

I like the first method (above) Here's a second method.

$\sin \left(3 x\right) = \sin \left(x + 2 x\right) = \sin x \cos \left(2 x\right) + \cos x \sin \left(2 x\right)$

$= \sin x \left({\cos}^{2} x - {\sin}^{2} x\right) + \cos x \left(2 \sin x \cos x\right)$

$= \sin x \left(3 {\cos}^{2} x - {\sin}^{2} x\right)$

So

${\lim}_{x \rightarrow 0} \sin \frac{3 x}{x} = {\lim}_{x \rightarrow 0} \frac{\sin x \left(3 {\cos}^{2} x - {\sin}^{2} x\right)}{x}$

$= {\lim}_{x \rightarrow 0} \left(\left(\sin \frac{x}{x}\right) \left(3 {\cos}^{2} x - {\sin}^{2} x\right)\right)$

$= \left(1\right) \left(3 {\left(1\right)}^{2} - {0}^{2}\right) = 3$