Question

Evaluate the limit below involving absolute value:
lim x→1+
(x^2 + 4x − 5) / (|x − x^2|)

Answer 1

Factor
\[\rightarrow \frac{(x-1)(x+5)}{|x(1-x)|}\]
then split absolute value into 2 cases...since you are evaluating at 1 from right
x is positive and (1-x) will be neg
\[\rightarrow \frac{(x-1)(x+5)}{-x(1-x)} = \frac{(x-1)(x+5)}{x(x-1)} = \frac{x+5}{x}\]
evaluate at 1
\[\lim = 6\]

Answer 2

since i'm evaluating 1 from the right, why is denominator -x(1-x) instead of (x(1-x))

Answer 3

because (1-x) is neg when x>1 ....the abs value makes it pos, or in other words multiply by -1

Answer 4

its a way to get rid of the abs value signs without affecting the value

Answer 5

whenever i have an absolute, do i always factor and focus on the inside parentheses

Answer 6

in situations like this?

Answer 7

yes...you have to pay attention to what values of x you are using and if it makes the inside "neg" or not

Answer 8

if its all pos values then drop the abs value signs and don't change anything

Answer 9

oh ok. now i see. thank you so much!

Answer 10

yw ...for practice try to see what the limit is if you evaluate from the left

Answer 11

hmm.. this doesn't seem right but i came up with 6. from the left, it should be negative. so when x<1 the x(x-1) should do

Answer 12

you are right, it should be -6

lets see...when x<1 , (1-x) is positive so dont change anything
\[\frac{(x-1)(x+5)}{x(1-x)} = \frac{(x-1)(x+5)}{-x(x-1)} = \frac{x+5}{-x} = -6\]
this time you factor out a neg, so you can cancel the (x-1)

Answer 13

why is it (x-1)? wouldn;'t it be negative. wouldn't the denominator be -x(1-x)

Answer 14

no but then you have exactly what we had before

the original abs value was |x(1-x)| , the inside can be either + or - depending on x right

so if x(1-x) is "-" then put a neg sign in front to make it pos
if x(1-x) is "+" then leave it alone

Answer 15

could it be x(1-x)

Answer 16

you mean could
\[|x(1-x)| = x(1-x)\]
?
yes if 0<x<1

Answer 17

oh i thought it could work for 1- since the inside of absolute value is negative

Answer 18

hmm 1- means approach from left ... 0.99 or something like that right
by plugging in this value for x ...(1-.99) = .01 which is positive

Answer 19

oh i get it

Answer 20

awesome :)
sorry i couldn't explain it better...absolute value can really throw people off

Answer 21

so everythign in absolute must be positive regardless of 1-. you have to change it so it becomes positive. i was thinking 1- would be negative overall and 1+ would be positive

Answer 22

no not at all , the plus and minus have nothing to do with sign , just whether you are approaching from right or left

Answer 23

obviously when x=1 then 1-x = 0
but what if x is 0.99 or 1.01 then you look at whether (1-x) is pos or neg

Answer 24

for 1- i was using numbers such as -2. that would change the outcome. so how did you determine it was from 0<x<1

Answer 25

well you want to pick a number that is really close to 1 since limit means approaching a number
thats why i chose 0.99 for 1-
and 1.01 for 1+

Answer 26

gotcha! i will definately keep that in mind. thanks again!