Question

Pic. help with b and c

Answer 1

Just b and c:)

Answer 2

so... I gather you have a table for \( \bf \large lim_{x \to -3^{-1}} g(x) \quad and \quad \lim_{x \to -3^{+1}} g(x)\)

Answer 3

yes i made a table and the limit of f(x) as x approaches -3 does not exist

Answer 4

but i cant tell what the limit is of f(x) as x approaches -3- and -3+

Answer 5

right, what can't you tell what the limit is for -3?

Answer 6

cause when I did my table I am confused which side is for -3- and which side is for -3+... let me show you

Answer 7

hhmmm \(\bf -3^{-1}\) means FROM THE LEFT, and \(\bf -3^{+1}\) means FROM THE RIGHT

Answer 8

so the values to the left of -3, will be for \(\bf -3^{-1}\)
and the values to the right of -3 will be for \(\bf -3^{+1}\)

Answer 9

so what would the limit be for −3−1?

Answer 10

as you notice there's quite a big jump between -3.001 and -2.999
it goes from 5001 to -4999
from way above, to way below

so for a limit to exist between both points, it has to have "continuity"
that is, the points have to display a continuous behaviour, so the jump has to be very slight in between

now for the limit to exist, "x" doesn't have to be = -3
just the close in the proximity on both sides of -3, have to be close enough to each other
in the case of \(\bf lim_{x \to -3^{-1}} g(x) \quad and \quad lim_{x \to -3^{+1}} g(x)\) is not the case
so the limit from both sides, as opposed to the one-sided limit
doesn't exist

Answer 11

yes i understand:) but idk what the limit of −3- is or for -3+

Answer 12

ohhh, ok.... well, that was for C .... right, I see you mean B

Answer 13

yes i mean b

Answer 14

I understand c:)

Answer 15

ok, I made a bigger table for -3, and it shows that as you go further away from -3, it seem to approach 0, and as you to -3, it seems to go to \(\bf +\infty\)

same behaviour on the right side, as it goes close to -3 it goes to \(\bf -\infty\)
as it goes away from it, goes down in value


so \(\bf \large lim_{x \to -3^{-1}} g(x) = + \infty \quad and \quad lim_{x \to -3^{+1}} g(x) = -\infty\)

Answer 16

oh okay thanks:)

Answer 17

notice that g(x) is a rational, and rationals have asymptotes
-2 and -3 are the roots of the denominator, and thus the vertical asymptotes
and usually at the asymptotes, you get that behaviour

Answer 18

how come the limit of -2 exists but not the limit of -3?

Answer 19

because -2 one-sided limits do have "continuity", or continuous behaviour

Answer 20

oh okay thank you:)

Answer 21

yw