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

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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
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@em2000 what did you get?
3) b 4) c

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Other answers:

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

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