Question

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

Answer 1

@welshfella

Answer 2

@em2000 what did you get?

Answer 3

3) b
4) c

Answer 4

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

Answer 5

Why not direct sub as it tells you to?

Answer 6

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

Answer 7

You should be able to do the rest

Answer 8

Yeah as he said

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

Answer 9

oh ok so 3) is c

Answer 10

Yeah, and what is the limit for 4?

Answer 11

-10

Answer 12

As in what is x approaching you have not put that

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

you cannot divide by zero in mathematics

Answer 17

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

Answer 18

so it does not exist

Answer 19

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

Answer 20

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

Answer 21

Astrophysics showed you the factored form of the top

Answer 22

what did u get>

Answer 23

-20

Answer 24

Yes.

Answer 25

sry bout that

Answer 26

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

Answer 27

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

Answer 28

Show your work and explain why

Answer 29

wait its 0

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

what

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

See example 1 and maybe read through his notes

http://tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx