# Thread: [SOLVED] logic for determining limits with absolute value.

1. ## [SOLVED] logic for determining limits with absolute value.

Given
$\displaystyle f(x)=\frac{x^2-1}{|x-1|}$

$\displaystyle \lim_{x\to1+}{f(x)}$

$\displaystyle \lim_{x\to1-}{f(x)}$

Is the numerator reducible? I am just not seeing the method I should start with, a push in the correct direction would be appreciated.

2. Originally Posted by hechz
Given
$\displaystyle f(x)=\frac{x^2-1}{|x-1|}$

$\displaystyle \lim_{x\to1+}{f(x)}$

$\displaystyle \lim_{x\to1-}{f(x)}$

Is the numerator reducible? I am just not seeing the method I should start with, a push in the correct direction would be appreciated.
$\displaystyle \frac{x^2-1}{|x-1|} = \frac{(x+1)(x-1)}{|x-1|}$

recall the definition of absolute value.

for $\displaystyle x < 1$ ...

$\displaystyle \frac{(x+1)(x-1)}{|x-1|} = \frac{(x+1)(x-1)}{-(x-1)} = -(x+1)$

for $\displaystyle x > 1$ ...

$\displaystyle \frac{(x+1)(x-1)}{|x-1|} = \frac{(x+1)(x-1)}{x-1} = x+1$

3. Thanks... I have no idea why I didn't see such a basic identity.

/goes to re-read HS algebra book