Question

find the indicated one-sided limit, if it exists:

lim x --> 2 (from the left side) of (x-3)/(x+2)

Answer 1

-1/4

Answer 2

how did you solve that?

Answer 3

i found the limit just by pluggin' in 2
since f is continous at x=2

Answer 4

plug in 2 for x

Answer 5

the right side limit is the same, btw

Answer 6

right!

Answer 7

it would be a different story if it was x-2 on the bottom

Answer 8

i think i get what you're saying hang on...

Answer 9

so if it asked x--> - 2 (from the left side) then what would the answer be?

Answer 10

would it be negative infinity?

Answer 11

yes

Answer 12

L'hopital's rule does not apply, since both the top and bottom are not indeterminate (the top is -1)

Answer 13

so we know f does not exist at x=-2
so a simple substittion will not work
and also we know if would approach one of the infinities
so if we plug in a number before -2 but close to -2 from the left
like -2.01
we get (-2.01-3)/(-2.01+2)
we get negative on top and negative on bottom
a negative divided by a negative is postive
so positive infinity would be the limit

Answer 14

you're right. damn, good call

Answer 15

yes good call!!!!:)))

Answer 16

thank you both!!

Answer 17

okay so i was confused about the L'hopital's rule thing

Answer 18

its ok, you'll learn it later in the quarter

Answer 19

from the right of -2 we can plug in -1.99 to determine what are infinity will be
(-1.99-3)/(-1.99+2)
so op top we get negative and on bottom we get positive
so the limit would be negative infinity for x approaching -2 from the right

Answer 20

it has to do with taking the limit if say both the top and bottom are zero or infinity

Answer 21

okay so as long as the number x is approaching (in this case it was 2) is defined in the function's domain, you can just plug it in? and if is NOT defined, THEN you would use the L'hopital thing?

Answer 22

use L'Hopital only if both the top and bottom are undefined. but if either the top or bottom is defined, then L'Hopital's rule cannot be used. however, if both the top and bottom are defined, then you just plug in. like your first one, as x approaches 2 from the left side, the function approaches -1/4. does that help?

Answer 23

yes, it makes sense...it was exactly what myininaya was talking about that i was originally trying to ask about....i wasn't sure (and still am a little confused) when you solve it that way? is it only if the number x is approaching isn't defined?

Answer 24

wait, solve it which way exactly?

Answer 25

myininaya Group Superhero
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so we know f does not exist at x=-2 so a simple substittion will not work and also we know if would approach one of the infinities so if we plug in a number before -2 but close to -2 from the left like -2.01 we get (-2.01-3)/(-2.01+2) we get negative on top and negative on bottom a negative divided by a negative is postive so positive infinity would be the limit

Answer 26

ok, i get what you are asking now. you already knew it was undefined at that point because 2-2 is obviously 0 and you can't divide by zero. because you are taking the limit from the left side at -2, the graph is increasing a very quick pace to positive infinity because it is a negative over a negative. So, in a very long way, yes this is the way to answer this question if its undefined. by the way, using the same method, the right side limit is negative infinity.

Answer 27

golly, I hope that helped.

Answer 28

it did!!!! thank you sooo much....i have ONE more question and hopefully this is a little simpler for you to explain....so back to the ORIGINAL problem: lim x --> 2 (from the left side) of (x-3)/(x+2) what would the solution be if it was approach 2 from the right side? same answer of -1/4?

Answer 29

the exact same. if you get confused, you can always graph it with a graphing calculator.

Answer 30

okay it just totally threw me off that the book asked from only one side, when the answer is the same for both! sorry to be so difficult, but you guys have clarified a lot for me! thank you especially, corey_flynn, for your time!

Answer 31

no problem, any time