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
$$\lim_{t\to 0}{\sin{\pi t}\over t}=\pi$$
Notice that for $\omega\neq 0$ we have \begin{align} \lim_{t\to 0}\frac{\sin \omega t}{t}&=\omega\lim_{t\to 0}\frac{\sin \omega t}{\omega t}\\ &=\omega\lim_{x\to 0}\frac{\sin x}{x}\qquad\text{Where }\,\,\,x=\omega t\\ &=\omega(1)\\ &=\omega \end{align}
Then $$\lim_{t\to 0}\left(-\frac{\sin \pi t}{t}\right)=\ldots?$$
You cannot factor $\pi$ out but you are close. You can do a change of variables. Assume that $u=\pi t$ (equivalently $t=u/\pi$). So as $t\to0$ we also have $u\to0$. Then $$\lim_{t\to0}\frac{-\sin(\pi t)}{t}=\lim_{u\to0}\frac{-\sin(u)}{u/\pi}=(-\pi)\lim_{u\to0}\frac{\sin(u)}{u}=-\pi\cdot 1=-\pi.$$
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.
