Question

limit as x approaches zero of quantity negative six plus x divided by x to the fourth power.

Answer 1

@Vocaloid

Answer 2

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{6+x}{x^4}}\)

this?

Answer 3

yes!

Answer 4

actually negative 6

Answer 5

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{6+x}{x^4}}\)

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{6}{x^4}+\lim_{x \rightarrow ~0}\frac{x}{x^4}}\)

so I don't think you will get anything defined out of the limit

Answer 6

wait, approaches -6?

Answer 7

oh, -6+x!
Doesn't matter

Answer 8

you would be then getting

\(\large \displaystyle \lim_{x \rightarrow ~0}\frac{-6}{x^4}+\frac{x}{x^4}\)

and still DNE

Answer 9

oh okay thanks!

Answer 10

Wait so which one is it?

Answer 11

it is infinity not zero, i think

Answer 12

I don't believe L'Hopital's rule applies, -6/0 doesn't count as an indeterminate form

Answer 13

no she meant part 2, but there x's xcancel

Answer 14

What is that rule though?

Answer 15

that is to differentiate top and bottom, IF
you get 0/0 or ∞/∞,
when you plug in the value that x approaches into the limit

Answer 16

this is L'Hospital's Rule

Answer 17

Could you guys help me with this?

Answer 18

So for example, I have:

\(\large \displaystyle \lim_{x \rightarrow ~0}\frac{\sin(x)}{x}\)

and there you would apply this rule
(can you tell me why?)

Answer 19

i don't see a reason for one-sided limit not to exist
(unless the function is totally not on that interval, or if it goes into infinity - asymptote)

Answer 20

Well x=2 is an asymptote

Answer 21

@phunish well, it's asking for the limit as the function approaches x = 2 from the left, any ideas?

Answer 22

ok, you see that the graph has two parts (two sticks :D)

right/

Answer 23

?