anonymous
  • anonymous
Pic. help with b and c
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
Just b and c:)
jdoe0001
  • jdoe0001
so... I gather you have a table for \( \bf \large lim_{x \to -3^{-1}} g(x) \quad and \quad \lim_{x \to -3^{+1}} g(x)\)

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anonymous
  • anonymous
yes i made a table and the limit of f(x) as x approaches -3 does not exist
anonymous
  • anonymous
but i cant tell what the limit is of f(x) as x approaches -3- and -3+
jdoe0001
  • jdoe0001
right, what can't you tell what the limit is for -3?
anonymous
  • anonymous
cause when I did my table I am confused which side is for -3- and which side is for -3+... let me show you
jdoe0001
  • jdoe0001
hhmmm \(\bf -3^{-1}\) means FROM THE LEFT, and \(\bf -3^{+1}\) means FROM THE RIGHT
jdoe0001
  • jdoe0001
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 \(\bf -3^{+1}\)
anonymous
  • anonymous
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anonymous
  • anonymous
so what would the limit be for −3−1?
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
yes i understand:) but idk what the limit of −3- is or for -3+
jdoe0001
  • jdoe0001
ohhh, ok.... well, that was for C .... right, I see you mean B
anonymous
  • anonymous
yes i mean b
anonymous
  • anonymous
I understand c:)
jdoe0001
  • jdoe0001
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 \(\bf \large lim_{x \to -3^{-1}} g(x) = + \infty \quad and \quad lim_{x \to -3^{+1}} g(x) = -\infty\)
anonymous
  • anonymous
oh okay thanks:)
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
how come the limit of -2 exists but not the limit of -3?
jdoe0001
  • jdoe0001
because -2 one-sided limits do have "continuity", or continuous behaviour
anonymous
  • anonymous
oh okay thank you:)
jdoe0001
  • jdoe0001
yw

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