Question

Let f(x) = (1-2x+x^2)/(abs(1-x))

find lim x--> 1+ and lim x-->1-

Answer 1

factor top
remeber
|1-x|=1-x if 1-x>0=>1>x
|1-x|=-(1-x)= if 1-x<0=> 1<x

Answer 2

after all that please tell me i didn't mess it up!

Answer 3

oh yes i did
!!!!!

Answer 4

after all that ... i must be tired. numerator is
\[(1-x)^2\] not
\[(1-x)(1+x)\]

Answer 5

is the answer 1 for both?
thats what i got

Answer 6

no

Answer 7

can you help zarkon?

Answer 8

\[\frac{(1-x)^2}{|1-x|}\]
\[\frac{1-x}{|1-x|}(1-x)\]
taking limit we get -1*0=0 or 1*0=0

Answer 9

it is still true that
\[\frac{1-x}{ |1-x|} = \left\{\begin{array}{rcc}
1 & \text{if} & x <1 \\
-1& \text{if} & x >1
\end{array}
\right.
\]
so you have
\[f(x) =\frac{(1-x)(1-x)}{|1-x|}= \left\{\begin{array}{rcc}
1-x & \text{if} & x <1 \\
x-1& \text{if} & x>1
\end{array}
\right.
\]

Answer 10

so (1-x) cancle with denominator and left with numerator 1-x right

Answer 11

is it continuous at x = 1?

Answer 12

no..it is not defined at 1

Answer 13

thanks