Question

Evaluate the limits
the lim of abs(2x-4)/2-x as x goes to 2+ , 2- and 2

Answer 1

so we have
\[ \Large \lim_{x\to2+}\frac{|2x-4|}{2-x} \]
right?

Answer 2

right

Answer 3

OK.
Since
\[ x\to2+ \]
then x>2 then
\[ \Large x>2 \]
\[ \Large 2x>4 \]
\[ \Large 2x-4>4-4=0 \]
then
\[ \Large |2x-4|=2x-4 \]
do you get this?

Answer 4

yes

Answer 5

OK, then
\[ \Large \lim_{x\to2+}\frac{|2x-4|}{2-x}=
\lim_{x\to2+}\frac{2x-4}{2-x}=\lim_{x\to2+}\frac{2(x-2)}{2-x}=-2 \]
right?

Answer 6

right

Answer 7

you can the same with x to 2- using
\[ \Large x\to2-\Rightarrow x<2 \]
tell me what do you get

Answer 8

ok the 1 you did at the top what did it come out to be. would the second 1 be -(2x-4) after you take the absolute value away.

Answer 9

Yes!
\[ \Large x<2 \]
\[ \Large 2x<4 \]
\[ \Large 2x-4<0\Rightarrow |2x-4|=-(2x-4) \]

Answer 10

what is the value of the limit on 2-

Answer 11

is it -2

Answer 12

are you sure?

Answer 13

I think so but not really sure

Answer 14

let's see
\[ \Large \lim_{x\to2-}\frac{|2x-4|}{2-x}=\lim_{x\to2-}\frac{-(2x-4)}{2-x}=
\lim_{x\to2-}\frac{-2(x-2)}{2-x} \]
\[ \Large =\lim_{x\to2-}\frac{-2}{-1}=2 \]

Answer 15

O ok so you canceled out the x-2 and the 2-x on top and bottom to get -2 over -1 Right ?

Answer 16

Yes!

Answer 17

now with these two calculations, your final queation:
does the limit
\[ \Large \lim_{x\to2}\frac{|2x-4|}{2-x} \]
exist?

Answer 18

no Right ?

Answer 19

right!

Answer 20

thanks you are great

Answer 21

you are welcome