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?
Best Answer
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.