Limit as t approaches 0 of $-\sin(\pi t)/(t)$ Sorry I am a highschool student who isn't too good at math. 

I know that the limit $\sin(t)/(t)$ as $t$ approaches 0 is 1. But how in the world would I evaluate

$$-\sin(πt)/(t)$$ 

as $t$ approches 0. I have not yet L'Hospital's rule yet so would there be an approach? Is it possible to factor out π from sin(πx)?
Note that 

${\sin{\pi t}\over t}=\pi{\sin{\pi t}\over \pi t}$

If we nos set $u=\pi t$ we get

$${\sin{\pi t}\over t}=\pi{\sin{u}\over u}$$

And $u\to 0$ when $t\to 0$ so ${\sin{u}\over u}\to 1$. We therefore have

$$\lim_{t\to 0}{\sin{\pi t}\over t}=\pi$$
If you have learned about Taylor expansions, you could also Taylor expand about t = 0 to find the limit. Like so :

$\sin {\pi t} \approx (\pi t) - \frac{(\pi t)^3}{6} + ...$

Substituting this in to your original expression and only taking the first term in the above approximation,

$\lim_{t\rightarrow0}-\frac{\sin {\pi t}}{t} \approx \lim_{t\rightarrow0}-\frac{\left(\pi t\right)}{t} = -\pi$.

Where the last step just involves cancelling out the $t$s.