Question

lim x-2+=((x+3)((x+2)/x(x+2))

Answer 1

cancel the common factor there that you ahve on top and bottom

Answer 2

the first x+2 is an absolute value will this make a difference? I didnt see the equation options before. should read
lim x 2(+)=\[\left( x+3 \right)*(\left| x+2 \right|/x+2)\]

Answer 3

answer should equal 1

Answer 4

does make a difference

Answer 5

as i suspected, didn't work canceling

Answer 6

\[\lim_{x \rightarrow 2^+}\frac{(x+3)|x+2|}{x(x+2)}\]
is this right?

Answer 7

just plug in 2 because the bottom isn't going to be zero

Answer 8

no the x+3 is before the equation of the \[\frac{ \left| x+2 \right|}{ x+2}\]

Answer 9

or did you mean -2^+

Answer 10

yes -2

Answer 11

\[\lim_{x \rightarrow -2^+}(x+3)\frac{|x+2|}{x+2}\]
so there is no x on bottom I guess?

anyways we have x approaches -2 from the right
that means x is greater than -2
x>-2
adding 2 on both sides gives
x+2>0
so since x+2>0 then |x+2|=x+2
so you can actually cancel the x+2's out (since we are approaching 2 from right)

Answer 12

-2 from right*

Answer 13

that makes sense but it still doesn't get the answer 1 thouhh

Answer 14

yes

Answer 15

it does

Answer 16

-2+3 is indeed 1

Answer 17

oh yes thank you, sometimes the easiest stuff hangs me up.

Answer 18

\[\lim_{x \rightarrow -2^-}(x+3)\frac{|x+2|}{x+2} \]
if we had approaches -2 from left
then we know x<-2
adding 2 on both sides would give us
x+2<0
so since we know x+2<0 then |x+2|=-(x+2)

Answer 19

so here you would have (-2+3)(-1)

Answer 20

anyways since right limit=1 and left limit=-1
then the actual limit wouldn't exist

Answer 21

just extra work if you wanted more examples

Answer 22

i appreciate the help, that does make sense there is no actual limit for the whole equation even though you can figure the left and right sides.

Answer 23

right :)
if the left=right then the actual would equal whatever that was they equaled :)