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I'm having trouble solving this limit:

$\lim_{x \to -2^-} \frac{1}{(x + 2)^2}$

I can't find a way to rationalize the denominator. Also, is there a way to do it without plugging in -2.001 and stuff or graphing it?

EDIT:

I realized after asking this question that it doesn't matter if you take the above limit from the right, left, or both. It's always $+\infty$. Here's an equation that gives $+\infty$ from the right and $-\infty$ from the left: $\lim_{x \to 3^+} \frac{x - 4}{x - 3}$

How do I (algebraically) determine if it is positive or negative?

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    You can think of it graphically as "move the graph two units to the right to position the singularity at the origin"...2011-05-16

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Maybe this way of thinking about it will seem a little more intuitive to you:

Let $\varepsilon > 0$, and consider the limit

$\lim_{x \rightarrow -2^{-}} \frac{1}{(x+2)^2} = \lim_{\varepsilon \rightarrow 0} \frac{1}{((-2-\varepsilon)+2)^2} = \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon^2}.$

Now the expression $\frac{1}{\varepsilon^2}$ can be made arbitrarily large by choosing $\varepsilon$ small enough, and so the limit does not exist.

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    This makes perfect sense! Thank you!2011-05-16
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You asked for an algebraic solution and that's well-covered in other answers, but I'd like to offer a conceptual way of reasoning through it. Let's start with your revised example: $\lim_{x \to 3^+} \frac{x - 4}{x - 3}$ As $x$ gets close to $3$, the numerator is getting close to $-1$ and the denominator is getting close to $0$. Something nonzero (rather, not tiny-approaching-zero) divided by something tiny-approaching-zero is going to be big (far from zero), suggesting a limit of $\infty$ or $-\infty$ (that is, suggesting that the limit does not exist, but may be of the specific $\infty$ or $-\infty$ kind of does-not-exist).

Now, how do we tell whether it's $\infty$, $-\infty$, or neither? Well, the numerator of the fraction is getting close to $-1$, so it's negative. The denominator of the fraction is getting close to zero, but specifically as $x\to 3^+$, $x>3$, so $x-3>0$ and the denominator is positive. The fraction is the quotient of a negative number and a positive number, so it's negative and $\lim_{x \to 3^+} \frac{x - 4}{x - 3}\to-\infty.$

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As $x \to -2^{-}$, your fraction becomes of form $\frac{1}{\epsilon}$, where $\epsilon$ is an arbitrarily small number, and 1 divided by a 'small' number is clearly a 'large' number (the smaller the denominator, the larger the value of the fraction). I hope you can see it from here.

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    @cable729: you consider the values: for every $x$ greater than $3$, $x-3$ is positive. For all $x$ between $3$ and $4$, $x-4$ is negative. So for $x$ sufficiently close to $3$, on the right, the fraction is negative (a negative number divided by a positive number).2011-05-16
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Given any $M > 0$, we have that $\forall x \in \left( -2 - \frac1{\sqrt{M}},-2 \right)$, we have $\frac1{(x+2)^2} > M$.

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    what it means is give me any positive number $M$, then if $x$ lies in the interval $\left( -2-\frac1{\sqrt{M}},-2 \right)$, then \frac1{(x+2)^2} > M. This implies that the fraction can be made arbitrarily large by getting "sufficiently close" to -$2$. $\forall$ stands for "forall" and $\in$ stands for "belongs to"2011-05-16