Question

Whoever can give me the right answer for all the questions gets a medal/fan! Look at picture

Answer 1

@Jaynator495 @JFraser @thomaster @UnkleRhaukus

Answer 2

\[\lim_{x\to0}\frac{-6+x}{x^4}\\
=\lim_{x\to0}\frac{-6}{x^4}+\lim_{x\to0}\frac{x}{x^4}\\=\]

Answer 3

-6?

Answer 4

\[\lim_{x\to0}\frac{-6+x}{x^4}\\
=\lim_{x\to0}\frac{-6}{x^4}+\lim_{x\to0}\frac{x}{x^4}\\
=-6\lim_{x\to0}\frac{1}{x^4}+\lim_{x\to0}\frac{1}{x^3}\\
=\]

Answer 5

what is \(\displaystyle\lim_{x\to0}\frac{1}{x^4}\)?

Answer 6

infinity

Answer 7

no

Answer 8

it is not defined
because
it could be infinity or -infinity

Answer 9

oh ok

Answer 10

so we get

\[\lim_{x\to0}\frac{-6+x}{x^4}\\
=\lim_{x\to0}\frac{-6}{x^4}+\lim_{x\to0}\frac{x}{x^4}\\
=-6\lim_{x\to0}\frac{1}{x^4}+\lim_{x\to0}\frac{1}{x^3}\\
=-6 \{\text{Not Defined}\}+\lim_{x\to0}\frac{1}{x^3}\\
= \{\text{Not Defined}\}\]

Answer 11

i.e. the limit dosen't exist

Answer 12

Got it?

Answer 13

yes thanks

Answer 14

And I need help with these 2 last ones:

Answer 15

good work you have #3, and #4 correct

Answer 16

i'm not sure about 10. or 11. but i don't think are correct

Answer 17

i can't remember all those tricky limit defintions

Answer 18

do you know 14 and 15

Answer 19

15 is not too bad

Answer 20

it's just like #3

Answer 21

\[s(t)=-2-6t\]

\[\dot s(t)=\frac{d}{dt}\Big(-2-6t\Big)\\
\qquad = \]

\[\dot s(2) = \]

Answer 22

understand ?

Answer 23

yes

Answer 24

Can someone help me with the 2 questions above?

Answer 25

I have no idea what this stuff is... I can't really help :/

Answer 26

Sorry, leme see if I can tag someone...

Answer 27

@dan815, can you try and help this person? Thnx!