Solving Absolute Value Limit: x→2

[Mr Davis 97]

Homework Statement

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

Homework Equations

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.

 

[Mark44]

Homework Statement

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

Homework Equations

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.

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

 

[Mr Davis 97]

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

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

 

[PeroK]

Homework Statement

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

Homework Equations

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.

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

 

[Mr Davis 97]

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

Pretty sure. Why?

 

[PeroK]

Pretty sure. Why?

See whether you can work it out!

 

[Mr Davis 97]

See whether you can work it out!

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

 

[PeroK]

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

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.

 

[HallsofIvy]

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".