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

- anonymous

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- anonymous

@welshfella

- Astrophysics

@em2000 what did you get?

- anonymous

3) b
4) c

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## More answers

- Astrophysics

\[\lim_{x \rightarrow 1} (x^2+3x-1) \implies ((1)^2+3(1)-1)\]

- Astrophysics

Why not direct sub as it tells you to?

- Astrophysics

\[(x^2-100) \implies (x-10)(x+10)\]

- Astrophysics

You should be able to do the rest

- Australopithecus

Yeah as he said
\[\lim_{x \rightarrow 1} x^2 + 3x - 1 = 1^2 + 3*1 - 1\]

- anonymous

oh ok so 3) is c

- Astrophysics

Yeah, and what is the limit for 4?

- anonymous

-10

- Astrophysics

As in what is x approaching you have not put that

- anonymous

ik it is lim x--> -10 ( i forgot to put the -10)

- Astrophysics

Oh ok, so you should be able to do an easy cancellation and your answer will be?

- Australopithecus

You need to use algebra to solve this because if you directly sub in 10 you will get a divide by zero

- Australopithecus

you cannot divide by zero in mathematics

- Astrophysics

\[\lim_{x \rightarrow -10} \frac{ (x-10)(x+10) }{ (x+10) }\]

- anonymous

so it does not exist

- Australopithecus

It exists, you just need to use algebra to solve it

- Astrophysics

Why do you say that? Do you see something that's getting cancelled

- Australopithecus

Astrophysics showed you the factored form of the top

- anonymous

what did u get>

- anonymous

-20

- Astrophysics

Yes.

- anonymous

sry bout that

- Australopithecus

Example:
\[\frac{(x+1)(x+2)}{(x+1)} = (x+2)*1\]

- 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

Show your work and explain why

- anonymous

wait its 0

- Astrophysics

Well that's not very useful, you're just guessing it seems, once you figure out the reason please do share!

- anonymous

(0^2 - 0x)/(0^4)= 0/0=0

- Astrophysics

No that's not how it works, try seeing if the limit exists by taking the right and left hand side limits.

- anonymous

what

- 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

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

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

See example 1 and maybe read through his notes
http://tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx

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