anonymous
  • anonymous
GIVING OUT MEDALS N FAN 3.)Find the limit of the function by using direct substitution. lim x-->1 (x^2+3x-1) a) 0 b) does not exist c) 3 d) -3 4.) Find the limit of the function algebraically lim x--> (x^2-100)/(x+10) a)-10 b)-20 c) does not exist d) 1
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@welshfella
Astrophysics
  • Astrophysics
@em2000 what did you get?
anonymous
  • anonymous
3) b 4) c

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Astrophysics
  • Astrophysics
\[\lim_{x \rightarrow 1} (x^2+3x-1) \implies ((1)^2+3(1)-1)\]
Astrophysics
  • Astrophysics
Why not direct sub as it tells you to?
Astrophysics
  • Astrophysics
\[(x^2-100) \implies (x-10)(x+10)\]
Astrophysics
  • Astrophysics
You should be able to do the rest
Australopithecus
  • Australopithecus
Yeah as he said \[\lim_{x \rightarrow 1} x^2 + 3x - 1 = 1^2 + 3*1 - 1\]
anonymous
  • anonymous
oh ok so 3) is c
Astrophysics
  • Astrophysics
Yeah, and what is the limit for 4?
anonymous
  • anonymous
-10
Astrophysics
  • Astrophysics
As in what is x approaching you have not put that
anonymous
  • anonymous
ik it is lim x--> -10 ( i forgot to put the -10)
Astrophysics
  • Astrophysics
Oh ok, so you should be able to do an easy cancellation and your answer will be?
Australopithecus
  • Australopithecus
You need to use algebra to solve this because if you directly sub in 10 you will get a divide by zero
Australopithecus
  • Australopithecus
you cannot divide by zero in mathematics
Astrophysics
  • Astrophysics
\[\lim_{x \rightarrow -10} \frac{ (x-10)(x+10) }{ (x+10) }\]
anonymous
  • anonymous
so it does not exist
Australopithecus
  • Australopithecus
It exists, you just need to use algebra to solve it
Astrophysics
  • Astrophysics
Why do you say that? Do you see something that's getting cancelled
Australopithecus
  • Australopithecus
Astrophysics showed you the factored form of the top
anonymous
  • anonymous
what did u get>
anonymous
  • anonymous
-20
Astrophysics
  • Astrophysics
Yes.
anonymous
  • anonymous
sry bout that
Australopithecus
  • Australopithecus
Example: \[\frac{(x+1)(x+2)}{(x+1)} = (x+2)*1\]
anonymous
  • anonymous
quick 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
  • Astrophysics
Show your work and explain why
anonymous
  • anonymous
wait its 0
Astrophysics
  • Astrophysics
Well that's not very useful, you're just guessing it seems, once you figure out the reason please do share!
anonymous
  • anonymous
(0^2 - 0x)/(0^4)= 0/0=0
Astrophysics
  • Astrophysics
No that's not how it works, try seeing if the limit exists by taking the right and left hand side limits.
anonymous
  • anonymous
what
Astrophysics
  • Astrophysics
Well 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
  • Australopithecus
You 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
  • Australopithecus
What 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
  • Australopithecus
See example 1 and maybe read through his notes http://tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx

Looking for something else?

Not the answer you are looking for? Search for more explanations.