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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$\begingroup$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.
$\endgroup$ 3 $\begingroup$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 $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}\to-\infty.$$
$\endgroup$ $\begingroup$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.
$\endgroup$ 5 $\begingroup$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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