Hi i got problem with solving number 47

My progress: i simplify to x^2-1

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- Feb 5th 2013, 03:23 AMPetrusLimits
Hi i got problem with solving number 47

My progress: i simplify to x^2-1 - Feb 5th 2013, 03:27 AMProve ItRe: Limits
Whenever you deal with absolute values, it helps to write it as its hybrid function form...

- Feb 5th 2013, 03:31 AMPetrusRe: Limits
What u mean with hybrid function form?

- Feb 5th 2013, 03:35 AMProve ItRe: Limits
$\displaystyle \displaystyle \begin{align*} |X| = \begin{cases} \phantom{-}X \textrm{ if } X \geq 0 \\ -X \textrm{ if } X < 0 \end{cases} \end{align*}$

- Feb 5th 2013, 03:51 AMPetrusRe: Limits
lx-1l x if>_1 and -x if x<1 i cant se a point

- Feb 5th 2013, 05:03 AMProve ItRe: Limits
So approaching from the left (where $\displaystyle \displaystyle \begin{align*} x < 1 \end{align*}$) your function will be $\displaystyle \displaystyle \begin{align*} \frac{x^2 - 1}{-(x - 1)} \end{align*}$. 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? - Feb 5th 2013, 05:19 AMPetrusRe: Limits
(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?

- Feb 5th 2013, 06:24 AMProve ItRe: Limits
Correct :)