I found the result to be $1/3\pi$ but with L'hopital. How can we solve this without L'hopital? $$ \lim_{ x \to \pi} \frac{ x\tan(x)}{x^3 - \pi^3}$$

You won't hardly get around standard limits. Here is a way using the standard limits $\lim_{u\to 0}\frac{\sin u}{u} = 1$ and $\lim_{u\to 0}\frac{ln(1+t)}{t}=1$: Using $\sin(\pi +u)=-\sin u$ and $x=1+t$ and considering $t\to 0^+$ you get \begin{eqnarray*}\frac{\sin^2\pi x}{x\ln^2x} & \stackrel{x=1...

is $\frac{8}{7}$ a rational number? my understanding is a rational number is one that terminates or repeats. However if you divide 7 by 8 you get a number that does not terminate or repeat. So am I right that this would not be a rational number then?