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anonymous
 one year ago
GIVING OUT MEDALS N FAN
3.)Find the limit of the function by using direct substitution.
lim x>1 (x^2+3x1)
a) 0
b) does not exist
c) 3
d) 3
4.) Find the limit of the function algebraically
lim x> (x^2100)/(x+10)
a)10
b)20
c) does not exist
d) 1
anonymous
 one year ago
GIVING OUT MEDALS N FAN 3.)Find the limit of the function by using direct substitution. lim x>1 (x^2+3x1) a) 0 b) does not exist c) 3 d) 3 4.) Find the limit of the function algebraically lim x> (x^2100)/(x+10) a)10 b)20 c) does not exist d) 1

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1@em2000 what did you get?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow 1} (x^2+3x1) \implies ((1)^2+3(1)1)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Why not direct sub as it tells you to?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[(x^2100) \implies (x10)(x+10)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1You should be able to do the rest

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Yeah as he said \[\lim_{x \rightarrow 1} x^2 + 3x  1 = 1^2 + 3*1  1\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, and what is the limit for 4?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1As in what is x approaching you have not put that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ik it is lim x> 10 ( i forgot to put the 10)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh ok, so you should be able to do an easy cancellation and your answer will be?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2You need to use algebra to solve this because if you directly sub in 10 you will get a divide by zero

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you cannot divide by zero in mathematics

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow 10} \frac{ (x10)(x+10) }{ (x+10) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it does not exist

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2It exists, you just need to use algebra to solve it

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Why do you say that? Do you see something that's getting cancelled

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Astrophysics showed you the factored form of the top

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Example: \[\frac{(x+1)(x+2)}{(x+1)} = (x+2)*1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0quick one, for this i got does not exist is that right? Find the limit of the function algebraically. lim x>0 (x^2  2x)/(x^4) a) does not exist b) 8 c) 0 d) 8

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Show your work and explain why

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Well that's not very useful, you're just guessing it seems, once you figure out the reason please do share!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(0^2  0x)/(0^4)= 0/0=0

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1No that's not how it works, try seeing if the limit exists by taking the right and left hand side limits.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Well I don't have the time to teach the concepts of limits but I suggest you watch this video and others related https://www.youtube.com/watch?v=riXcZT2ICjA

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2You cannot divide by zero, so if you see zero as the denominator after doing a substitution you need to either solve algebraically or test both sides of the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2What I mean by both sides is that you take a limit using an extremely small number from the left we will denote it: \[0^\] This would be like 0.00000000000000000000006 or something so small its almost zero but not quite you do the same for the other side (right side) \[0^+\] This will be 0.0000000000000000000000003 or something so small its almost zero but not quite If \[\lim_{x \rightarrow 0^} f(x) = \lim_{x \rightarrow 0^+} f(x)\] The limit is said to exist. If they do not equal the limit does not exist Note that being said, when a number is extremely tiny and in the denominator your function is going to equal infinity

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2See example 1 and maybe read through his notes http://tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx
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