# Thread: Value and limits of a function

1. ## Value and limits of a function

This problem appeared on the first test in differential calculus. Since it does not involve any calculus I am hoping this is the right place to present it.

For this problem, consider f(x) = / x + 2 / / (x + 2)
If f(x) is expressed without absolute value, f(x) = ?

EVERYONE including me, thinks it should be = 1 The professor says it is equal to 1 or -1.

Can someone please explain why? We are also suppose to determine the intervals where f(x) is continuous. Again, I do not understand why they are from negative infinity, - 2 union with (-2, infinity)

Any help would be much appreciated. Frostking

2. Originally Posted by Frostking
This problem appeared on the first test in differential calculus. Since it does not involve any calculus I am hoping this is the right place to present it.

For this problem, consider f(x) = / x + 2 / / (x + 2)
If f(x) is expressed without absolute value, f(x) = ?

EVERYONE including me, thinks it should be = 1 The professor says it is equal to 1 or -1.

Can someone please explain why? We are also suppose to determine the intervals where f(x) is continuous. Again, I do not understand why they are from negative infinity, - 2 union with (-2, infinity)

Any help would be much appreciated. Frostking
I assume this is actually

$f(x) = \frac{|x + 2|}{x + 2}$.

The wording of the question is very ambiguous. I assume that to "write this without the absolute value sign" means to write an equivalent expression without the absolute value sign - not just to remove the sign.

Remember that $|x| = x$ if $x < 0$ and $|x| = -x$ if $x \leq 0$.

So here, $|x + 2|$ can equal either $x + 2$ if $x + 2 \geq 0$ or $-(x + 2)$ if $x + 2 < 0$.

So Case 1: $|x + 2| = x + 2$.

$f(x) = \frac{x + 2}{x + 2} = 1$.

Case 2: $|x + 2| = -(x + 2)$

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