How to tell when a limit diverges?

Question

I have $\underset{x \to -2} \lim \frac{x^2-1}{2x+4}$ which doesn't exist because it diverges. I spent awhile trying to remove the division by zero before plugging it into an online math calculator, and I want to know how I could have known that it diverges. I'm not skilled enough to just look at it and figure out there's no way for me to reform it so it doesn't have zero division when $x=-2$. Is there a way for me to find out whether it diverges other than just working on it for awhile and then guessing that the limit diverges if I can't get rid of the zero division?

My Steps:

$$\underset{x \to -2} \lim \frac{x^2-1}{2x+4}$$

$$=\underset{x \to -2} \lim \frac{(x^2-1)(2x-4)}{2x+4(2x-4)}$$

$$=\underset{x \to -2} \lim \frac{(x^2-1)(2x-4)}{4x^2-16}$$

$$=\underset{x \to -2} \lim \frac{(x^2-1)(2x-4)}{4(x^2-4)}$$

$$=\underset{x \to -2} \lim \frac{(x-1)(x+1)2(x-2)}{4(x-2)(x+2)}$$

$$=\underset{x \to -2} \lim \frac{2(x-1)(x+1)}{4(x+2)}$$

$$=\underset{x \to -2} \lim \frac{(x-1)(x+1)}{2(x+2)}$$

$$=\underset{x \to -2} \lim \frac{x^2-1}{2x+4}$$

Answer 1

There's not much use in these algebraic manipulations of the fraction.

Since the numerator does not tend to $0$ for $x \to -2$, but the denominator does, the absolute value of the function (the fraction) will tend to infinity.

This is sufficient to conclude that there will not be a finite limit but you could still say the limit is either $+\infty$ or $-\infty$. However, we reserve this for the case where the left- and right-handed limits agree, so when they are both either $+\infty$ or $-\infty$ respectively.

Now in your case: $$\lim_{x \to 2^-}\frac{x^2}{2x+4} = \ldots$$ but $$\lim_{x \to 2^+}\frac{x^2}{2x+4} = \ldots$$ so...

Answer 2

$$ \frac{x^2}{2x+4}= \frac{x^2-4+4}{2(x+2)}= \frac{(x-2)(x+2)}{2(x+2)}+\frac{4}{2(x+2)}=\\ \frac{x-2}{2}+2\frac{1}{x+2}. $$

It's not difficult to see that as $x$ approaches $-2$ from the left, $\frac{1}{x+2}$ goes to negative infinity: $\lim_{x\to-2^-}\frac{1}{x+2}=-\infty$. And as $x$ approaches $-2$ from the right, $\frac{1}{x+2}$ goes to positive infinity: $\lim_{x\to-2^+}\frac{1}{x+2}=+\infty$. So, that limit clearly does not exist and by extension it must be the case that the entire limit does not exist either. In more precise mathematical language, it all looks like this:

$$ \lim_{x\to-2^-}\frac{x^2}{2x+4}=\lim_{x\to-2^-}\left(\frac{x-2}{2}+2\frac{1}{x+2}\right)=-2+2\cdot(-\infty)=-2-\infty=-\infty,\\ \lim_{x\to-2^+}\frac{x^2}{2x+4}=\lim_{x\to-2^-}\left(\frac{x-2}{2}+2\frac{1}{x+2}\right)=-2+2\cdot(+\infty)=-2+\infty=+\infty. $$

For a limit to exist, the two one-sided limits should be equal the same value. This is clearly not the case.