/sci/ - Science & Math

Let's take the function 1/x
Pic related.

Now let's start approaching zero from the right. The number gets larger and larger, until it goes all the way to infinity.
Ok, so start approaching zero from the other side. The number gets larger and larger in the negative direction, until it goes all the way to negative infinity.

So given that, what is 1/0? Is it infinity? Negative infinity? Somewhere in the middle?
Answer: it's undefined, because you can't divide by zero you shit.

>>9737377
What happens to f(x) as x approaches zero

>>9541880
No, you fucked it up.
6/(-3) = -2
That's POSITIVE SIX over NEGATIVE THREE.
6/(-2) = -3
6/(-1) = -6
6/(0) = -inf

I can count to 0 by 3 2 1 0 or by -3 -2 -1 0. These have different limits. Would 6/0 be defined by the first limit but not the second?

>>9541885
For any binary operation * and a real number r, r * inf is equal to the limit of r * x as x goes to infinity. Addition and multiplication are associative and commutative when defined.

>>7363776
If you tried to define 1/x as x approaches x, then you would get two answers at once, but -infinity and +infinity. You didn't take into account approaching zero from the negative numbers.

First off, 1 split into 1 pieces result in each piece being 1. 1/1=1

For the second part look at it this way.
Take the graph of 1/x. As x gets smaller, 1/x gets much bigger.
1/0.001 =1000
1/0.000001=1000000
So, we says that 1/x approaches infinity as x approaches 0.

idk, if that makes sense...

Why isn't the area under f(x)=1/x, x>0 considered strictly lesser than the area under f(x)=x^2?
Both are infinite but the former keeps getting thinner and thinner while the latter keeps growing and very quickly. Is there a way to express the difference between these infinite areas?

The graph of f(x)=1/x never touches f(x)=0 but does it touch f(x)=ɛ, where ɛ is infinitesimal?

Is the area under 1/x smaller than the area under x^2?
I think it's clearly smaller and it's asinine to consider them equal. But is there a way to actually define this?

Have you tried looking at a the graph of the function 1/x?

Look at the thing diverging.

<div class="math">\int \frac{1}{x} dx</div>
Integration by parts:
<div class="math">\int \frac{1}{x} dx = \left[\frac{1}{x} \cdot x - \int -\frac{x}{x^2} dx\right] = 1 + \int \frac{1}{x} dx</div>
Doesn't this imply 1 = 0? Where's the fallacy?

You know how f(x) = 1/x wraps around from negative infinity to positive infinity at zero? It works kind of like that.

>>6125377
Holy shit OP you just realized something everybody already knows!

>>5076616

You are silly.

It was already proven.
> In rhodium, 280 pK and -750 pK have been reached
> In rhodium, 280 pK and -750 pK have been reached
> In rhodium, 280 pK and -750 pK have been reached
> In rhodium, 280 pK and -750 pK have been reached
> In rhodium, 280 pK and -750 pK have been reached

Also, of course something can exist and not reach zero. Picture related. -1/x function. (consider x as temperature, if you please)

Why does 1/0 not equal infinity?

1/infitiy equals 0, which is quite intuitive. As the denomiator gets huge, the quotient becomes very small, ie zero.

But why shouldn't it work the other way around as well? As the denominator becomes very very tiny, the quotient becomes huge, ie infinity.

pic related, 1/x

ok im taking calculus and analytical geometry and im getting my ass kicked by it (currently getting a D) i need to get a C at least in this class any suggestions???

pick related (my grade if i dont fix this

The teacher likes to say that stuff will be on the test then not put it on the test which is a major problem. He also like to put tricks in his problems to mess you up.