# Thread: What's happening here? (Inequalities)

1. ## What's happening here? (Inequalities)

We are doing limits, that comes under Calculus I in most books, but it's part of Pre-calculus (In the Stewart book)

I know how limits work, my question is about their last few steps, they go from "x > -2" to "x^2 + x - 2 < 0" in just one line. I know this has to do with pre-calc but my professor assumes we already know the logic behind it, when I do not

Are there any intermediary steps? Or do you just take any number greater than (-2) and plug it in x^2 + x - 2?

2. ## Re: What's happening here? (Inequalities)

The problem, as you give it, just doesn't make sense. Is the "-2" that point at which the limit is taken? That is, should the first part be $\lim_{x\to -2}$? If so then the "+" doesn't make sense.

Ahh! Could it be $\lim_{x\to -2^+}$? If so then that means "take the limit as x goes to -2 from above".
That is, we only consider values of x that are larger than -2. Notice that the denominator, $x^2+ x- 2$, is 0 when x= -2. Since the numerator, 3, does not go to 0, the is no finite limit and the only question is whether the fraction goes to " $+\infty$" or " $-\infty$". Since the numerator of the fraction, 3, is positive, the sign of the limit depends only on the sign of the denominator, for x larger than -2 but close to -2. In particular, we know that because $x^2+ x- 2= 0$ for x= -2, x-(-2)= x+ 2 must be a factor. That makes it easy to see that $x^2+ x- 2= (x+ 2)(x- 1)$. For x< 1, x- 1 is negative (and since we are taking the limit as x approaches -2, any x of interest is less than 1). The denominator, and so the entire fraction, will be negative if and only if x+ 2 is positive. x+ 2> 0 give x> -2. As we approach x from above, x is always greater than -2 so x+ 2> 0 and then $(x+ 2)(x- 1)= x^2+ x- 2< 0$

3. ## Re: What's happening here? (Inequalities)

note that $\dfrac{3}{x^2+x-2} = \dfrac{3}{(x+2)(x-1)}$

as $x \to -2$ from the left (think about -2.01 or closer if you like) , the value of $(x+2)$ approaches a very small negative value. The numerator stays positive and the other factor in the denominator stays negative. So, what you end up with is ...

$\dfrac{\text{fixed positive value}}{(\text{small negative value})(\text{relatively fixed negative value})} \to \text{large positive value}$

as $x \to -2$ from the right (think about -1.99 or closer if you like) , the value of $(x+2)$ approaches a very small positive value. The numerator stays positive and the other factor in the denominator stays negative. So, what you end up with is ...

$\dfrac{\text{fixed positive value}}{(\text{small positive value})(\text{relatively fixed negative value})} \to \text{large negative value}$

therefore, $\displaystyle \lim_{x \to -2^+} \dfrac{3}{(x+2)(x-1)} = -\infty$