Question

more derivatives .-.

Answer 1

haha so now this one comes up

well apply what i wrote the first time

Answer 2

btw this isn't a derivative problem

Answer 3

well -6+0/0^4=-6/0=0 c: wait the what is it if not a derivative problem? .-.

Answer 4

no -6/0 = -infinity

cant divide by 0 .... as you divide my a smaller number the result gets infinitely bigger

Answer 5

lol its a limit problem

Answer 6

besides XD I told you i had more questions
and im not looking for answers such as infinity and -infinity.
im looking for an actual number and or the limit that it dosent extist

Answer 7

A derivative is a "rate of change" -- how does one variable change when another variable changes? In the average case, this is a secant; in the instantaneous case, this is a derivative.

Limits are used to derive the derivative, e.g., the derivative is defined as:

dy/dx = lim h->0 (f(x+h) - f(x))/h.

Answer 8

oh ok..... if divide by 0, limit does not exist

Answer 9

.-. wtf why wouldnt it be 0? cuz anything multiplied or divided by 0 is 0....

Answer 10

oh well.... another qustion that i dont get!

Answer 11

so if i plug in 2, i get 2^2+3(2)-1=4+6-1=9? right?

Answer 12

try putting 6/0 into your calculator

Answer 13

Dividing by 0 is not a valid operation, but when discussing limits, something like lim x->0 1/x^2 is +infinity, not 0.

Answer 14

Yes.

Answer 15

yes limit= 9 for that problem

most limits are easy substitutions, they only become tricky when dealing with infinity and indeterminate forms

Answer 16

wait why 6/0?

Answer 17

you can do it from L'Hospital rule. In this rule you just need to take derivative of nominator and dinominator. Thus, it equals to [d/dx(-6+x)]/[d/dx(x^4)]
Therefore it becomes lim x goes to 1/4x^3=1/0= plus infinity

Answer 18

what is said L'Hospital rule?

Answer 19

it is just a useful by applying derivative to limit. you just need to tak derivative of up and down as i did, but it cannot be applied for 0/infinity or infinity/0