How can I solve $ \displaystyle \lim_{x \to 3} \frac{2 - x}{(x - 3)^{2}} $ without using l’Hôpital’s Rule? I know that the limit is $ - \infty $, but I do not understand how to get to this solution without using l’Hôpital’s Rule.
 A: The expression
$$\frac{2-x}{(x-3)^2}$$
doesn't have an indeterminate form when $\,x\to 3\,$...the limit doesn't exist finitely, BTW.
If you really wanted
$$\frac{3-x}{(x-3)^2}=-\frac{x-3}{(x-3)^2}=-\frac{1}{x-3}$$
and again the limit doesn't exist...
A: L'Hopital's rule is not applicable to situations like this.  It is applicable to limits in which the numerator and denominator both approach $0$ or both approach $\infty$.  If the numerator approaches $0$ and the denominator approaches something else, then the limit is $0$.  If the denominator approaches $0$ and the numerator approaches something else, then the limit is $\infty$ (except that if you distinguish between $+\infty$ and $-\infty$ then some issues arise).
A: Imagine $x$ very close to $3$ but not equal to $3$. Then $2-x$ is very close to $-1$, and $(x-3)^2$ is a positive number very close to $0$. Thus, when we divide, we get a very large negative number. 
You could, by hand or with a calculator, compute $\dfrac{x-2}{(x-3)^2}$ for various values of $x$ close to $3$, such as $x=3.01$, $x=2.998$, $x=3.0003$, $x=2.999997$, to get an idea of what's happening.
Or else, if you have reliable graphing software, ask it to plot $y=\dfrac{x-2}{(x-3)^2}$, for values of $x$ in a reasonably narrow window around $x=3$.
A: Informally, we have
$$
\lim_{x \to 3} \frac{2 - x}{(x - 3)^{2}} = \frac{-1}{0^{+}} = - \infty,
$$
where we require the fact that $ (x - 3)^{2} \longrightarrow 0^{+} $ as $ x \longrightarrow 3 $. This proves, heuristically, that the limit is $ - \infty $ and not a finite number.
