Question

HELP: SKETCHING evaluating limits with vertical asymptotes.
INFINITY OR -INFINITY OR DOES NOT EXIST.
(ATTACHED BELOW)

Answer 1

@kropot72

Answer 2

they're all N. See the repsonse I sent previous to the other question you had.

Answer 3

i know i've read it over but i don't understand why they are ALL negative..... :/ this is so confusing. i really haven't struggled this much on a problem.

Answer 4

@pgpilot326

Answer 5

@asnaseer ! :)

Answer 6

please help.

Answer 7

sorry, i think i goofed.
let's look at the first one...

Answer 8

goofed? lol @pgpilot326

Answer 9

let's start with this one. it's complicating.

Answer 10

i plugged in -4 for the left side and for the right side, -2

Answer 11

see, the only way i know is by sketching, but even that is wrong.

Answer 12

As the denominator can be conveniently rewritten to \((x+3)^{2}\), it is clear enough that the denominator is ALWAYS positive. You must find the sign only of the numerator.

This thing is negative for all x such that

2x+4 < 0
or
2x < -4
or
x < -2

Where does that leave values around x = -3?

Answer 13

\[\lim_{x \rightarrow -3^{-}}\frac{ 2x+4 }{ x ^{2}+6x+9 }=\lim_{x \rightarrow -3^{-}}\frac{ 2x+4 }{\left( x+3 \right)^{2} }=\lim_{x \rightarrow -3^{+}}\frac{ 2x+4 }{\left( x+3 \right)^{2} }=\lim_{x \rightarrow -3}\frac{ 2x+4 }{\left( x+3 \right)^{2} }\]
because the bottom is squared, it doesn;t matter if the inside is positive or negative, the bottom will always be positive. the top will always be negative. since the bottom goes to 0, the whole thing will go to neg infinite for each case.

Answer 14

@tkhunny on the negative side.

Answer 15

so i see that for the denominator it will always be a positive. now the numerator is we = to zero it would be -2 right, so the limit is infinity?

Answer 16

-infinity**

Answer 17

right but i want to be able to see that by sketching the graph.

Answer 18

okay so let's look for the first one. -3^-
we know the number -2 but so what. why is it negative infinity.

Answer 19

i can't imagine this on the top of my head "oh negative" just because the negative 2 is there. how can i see that on the graph?

Answer 20

same applies to -3^+ ... "oh negative again" i want to be able to see it please.

Answer 21

Little known fact: The degree of the factor causing the asymptote controls its behavior.

Example: \((x+3)\) This must mean +\(\infty\) on one side and -\(\infty\) on the other side.

Example: \((x+3)^{3}\) This must mean +\(\infty\) on one side and -\(\infty\) on the other side.

Note how each has ODD degree. This is opposed to those of EVEN degree:

Example: \((x+3)^{2}\) This must mean either +\(\infty\) on both sides or -\(\infty\) on both sides.

Example: \((x+3)^{4}\) This must mean either +\(\infty\) on both sides or -\(\infty\) on both sides.

Thus, once we encounter an asymptote-causing factor of EVEN degree, it is of no consequence if the limit is from the right or the left.

Also note, with x < -2 being important, what does the right or left limit care? Always from the left and eventually from the right, the values are less than -2.

Answer 22

i already started the graph.