Limits

**Petrus** · February 5th 2013, 03:23 AM

Hi i got problem with solving number 47
My progress: i simplify to x^2-1

**Prove It** · February 5th 2013, 03:27 AM

Whenever you deal with absolute values, it helps to write it as its hybrid function form...

**Petrus** · February 5th 2013, 03:31 AM

What u mean with hybrid function form?

**Prove It** · February 5th 2013, 03:35 AM

$\displaystyle \begin{align*} |X| = \begin{cases} \phantom{-}X \textrm{ if } X \geq 0 \\ -X \textrm{ if } X < 0 \end{cases} \end{align*}$

**Petrus** · February 5th 2013, 03:51 AM

lx-1l x if>_1 and -x if x<1 i cant se a point

**Prove It** · February 5th 2013, 05:03 AM

So approaching from the left (where $\displaystyle \begin{align*} x < 1 \end{align*}$ ) your function will be $\displaystyle \begin{align*} \frac{x^2 - 1}{-(x - 1)} \end{align*}$ . Simplify and subsitute x = 1 to find the left hand limit.

What do you think you would have to do for the right hand limit?

**Petrus** · February 5th 2013, 05:19 AM

(X^2-1)/(x-1) with other words if we factor ((x-1)(x+1))/(x-1) so we got -2 from left and 2 for right? I am correct?

**Prove It** · February 5th 2013, 06:24 AM

Correct

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