Question

As the limit approaches 0 the limit is (e^x-1)/x^3

Answer 1

ok so you do have 0/0 when you plug in 0
what did you get after one around of l'hospital ?

Answer 2

I got e^x/3x^2

Answer 3

correct

Answer 4

and as you said in your direct message you got 1/0 when pluggin in 0

now e^x is always positive
3x^2 is always positive

so we know what kind of does not exist we have
we have positive infinity

Answer 5

Hmmmm

Answer 6

I'm confused

Answer 7

1/0 means the limit does not exist

Answer 8

Yes

Answer 9

there are a types of does not exist:
1) positive infinity( we do have positive over positive)
2) negative infinity (we don't have negative over positive or positive over negative)
3) jump (we don't have that type of thing here)
4) oscillating function ( I'm pretty sure functions can only oscillate between numbers if x is approaching one of the infinities)

Answer 10

for 3 and 4 we normally just limit does not exist

Answer 11

but 1 and 2 we can say the limit does not exist or be more specific if possible and say one of those infinities

Answer 12

Okay

Answer 13

I thought it would be 0 for some reason

Answer 14

if you had 0/1 yes the limit would be 0

Answer 15

wait ohhhhh I see now wow I feel silly

Answer 16

its cool
lots of people get division by 0 confused with 0 divided by some other number

Answer 17

Me lol

Answer 18

:)

Answer 19

:) thank you