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I know that the limit is $ - \infty $, but I do not understand how to get to this solution without using l’Hôpital’s Rule.

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4 Answers 4

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...

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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).

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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.

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The numerator is $2-x$ (in the limit -1). –  Américo Tavares Mar 4 '13 at 1:39
    
@AméricoTavares: Thanks for spotting the typo error! –  Haskell Curry Mar 4 '13 at 1:41
    
You are welcome! –  Américo Tavares Mar 4 '13 at 1:55

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 $x=3$.

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