Question

Find the limit of the function algebraically.

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

Answer 1

im here

Answer 2

sorry let me make it simpler limit as x approaches zero of (-6+ x)/x^4

Answer 3

I think the limit does not exist @geerky42

Answer 4

Well, it does approaches to \(-\infty\), I just can't see how to solve it algebraically.

Answer 5

how would you solve it then @geerky42

Answer 6

my options are
6
0
-6
does not exist

Answer 7

I know to solve algebraically you can solve by replacing x with 0 but that would give you 0 in the denominator so it cant be -6

Answer 8

Sorry, I'm struggling with it. But if we graph it, we can see that "Does not exist" would be best option.

@amistre64 @dan815 probably can explain how to solve it algebraically.

Answer 9

Okay well thank you @geerky42 :)

Answer 10

can you write the expression is math notation?

Answer 11

limit as x approaches zero of (-6+ x)/x^4

didnt see that one ...

Answer 12

algebra .... the limit of a function is defined only if the left and right limits exist, and are the same.

trial and error means that we make x as close to zero as we want ...

let x1 = .0000000001
let x2 = -.0000000001

are they almost the same? or are they light years apart?

Answer 13

They are almost the same

Answer 14

but how the same are they?

-6348398458973852375723067302974650723 is pretty big, and its only getting bigger

Answer 15

where did you get that number from?

Answer 16

i did exactly like a suggested you do ...


http://www.wolframalpha.com/input/?i=%28x-6%29%2Fx%5E4%2C+x%3D-.0000000001

Answer 17

-6 * 10^(40) is a pretty big number

Answer 18

and its not 6, -6 or 0

Answer 19

ohhhh okay I see.

Answer 20

So the limit does not exist?

Answer 21

in your course, they define it as not existing, in other courses they define it as negative infinity

we have to choose an option, sooo ..... dne is it

Answer 22

okay so it would exist if the left and right limit equal each other?

Answer 23

.. if the left and right limit are the same finite number, in your case

Answer 24

infinity is by default, not finite

Answer 25

Okay thank you @amistre64 :)

Answer 26

Do you mind helping me with another problem?

Answer 27

i spose one more is fine, depend son how long my sanity hold out :)

Answer 28

Haha i'll make it quick
Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1/(x-7)^2 as x approaches 7.

Answer 29

do you know what a removable discontinuity is?

Answer 30

I forgot

Answer 31

its common factors top to bottom, common factors limit to 1 so they can be removed

what left over defines bad xs.

does the setup simplify any?

Answer 32

this of course isnt graphing or tabling ... those are busy work that just take up way to much time and effort

Answer 33

do I have to do (x-7)(x-7)?

Answer 34

Sorry im still confused 0o0

Answer 35

x-7 = 0 regardless of if you split it into to parts

the top has no x-7 factor so we cant remove anything

(x-7)^2 is a number that is 0 or more regardless of x so that tells us that regardless of the direction we are heading, we are moving in a positive direction.along a vertical boundary that cannot be crossed.

Answer 36

oh, x-7 is bad when x-7=0, why?

Answer 37

Ummm is that because x=7

Answer 38

x=7 makes a bad math ... 1/0 is not good to get

Answer 39

spose we had:

3(x-7)
---------
(x-7)(x-3)

the limit as x to 7, well ... we can remove some parts and simplify

3/(x-3) , when x=7 is the limit since there are no bad maths left to cause issues

Answer 40

so x=7 is the vertical asymptote?

Answer 41

yep, its something that cannot be ignored/removed.

we cant just step over it and keep on moving, (we can ignore removable stuff cant we?)

Answer 42

removable stuff is like a crack in the sidewalk, we can step over it and carry on

a vertical asymptote is a brick wall, its cant be removed and it blocks our path. we have to find a way around it, we move up or down it in a vain attempt to get past it

Answer 43

Ohhh wow that's a really helpful comparison

Answer 44

Thank you soooooo much @amistre64!

Answer 45

your welcome, ill leave the tabling and graphing to you :)