# How do you find the limit of (sin(5x)) / (x-π) as x approaches pi?

Sep 23, 2016

I used the fundamental trigonometric limit ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$. I got a limit of $- 5$. (Which I did check using graphing technology.)

#### Explanation:

To use the fundamental limt, the argument =of the sine function must be the same as the denominator and must have a limit of $0$ adn $x \rightarrow \text{ whatever}$ (In this case, as $x \rightarrow \pi$.)

We can get a $5 x$ in the denominator by multiplying by $5$, but when we do that we get

$\sin \frac{5 x}{x - \pi} = 5 \sin \frac{5 x}{5 x - 5 \pi}$.

We need $\sin \left(5 x - 5 \pi\right)$ in the numerator.

Let's try using the difference formula for sine (if that doesn't work we'll try another formula or a different approach.)

sin(5x-5pi) = sin(5x)cos(5pi)-cos(5x)sin)5pi)

$= \sin \left(5 x\right) \left(- 1\right) - \cos \left(5 x\right) \left(0\right)$

$= - \sin \left(5 x\right)$.

We can conclude that $\sin \left(5 x\right) = - \sin \left(5 x - 5 \pi\right)$.

So we want

${\lim}_{x \rightarrow \pi} \sin \frac{5 x}{x - \pi} = {\lim}_{x \rightarrow \pi} \left[- 5 \cdot \sin \frac{5 x - 5 \pi}{5 x - 5 \pi}\right]$

$= - 5 {\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta}$

$= - 5 \left(1\right) = - 5$