# Absolute value limit

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1. Oct 23, 2015

### Mr Davis 97

1. The problem statement, all variables and given/known data
Solve: $\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4}$.

2. Relevant equations

3. The attempt at a solution
I am trying to solve the following limit: $\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4} = \lim_{x\rightarrow 2} \frac{\left | (x + 1)(x + 2) \right |}{(x - 2)(x + 2)}$. However, I am not sure how to proceed. Normally, I would cancel out the factors, but I am not sure what do with the absolute value in the numerator.

2. Oct 23, 2015

### Staff: Mentor

Since x is near 2, |x + 2| will be near 4; i.e., positive.

3. Oct 23, 2015

### Mr Davis 97

So I can just proceed as if there were no absolute value?

4. Oct 23, 2015

### PeroK

Are you sure it's not supposed to be $x^2 - 3x + 2$ in the numerator?

5. Oct 23, 2015

### Mr Davis 97

Pretty sure. Why?

6. Oct 23, 2015

### PeroK

See whether you can work it out!

7. Oct 23, 2015

### Mr Davis 97

Well I got DNE! Isn't that still a valid response?

8. Oct 23, 2015

### PeroK

For the original limit, assuming DNE means "does not exist", that's correct.

$\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 - 3x + 2 \right |}{x^2 - 4}$.

Is a better problem, because it's of the form $\frac{0}{0}$ which the original problem wasn't.

9. Oct 24, 2015

### HallsofIvy

Staff Emeritus
The point is that as x goes to 2, the denominator goes to 0 but the numerator does NOT. For x arbitrarily close to 0, the numerator close to 18, denominator close to 0, the fraction will be huge. If the limit were as x goes to -2, then it would be of the form "0/0".