Question

If f(x)=3x^2+7x then what's f(x)/x=? What is this when it is evaluated at the points x=1 and x=0?

Answer 1

3x^2+7x /x ?
mm i think at =1
then its 10
and at 0 its not continues but limit =7

Answer 2

3x+7 i guess
and at x=1 we have 10
and at x=0 we have 7 :D

Answer 3

Why do we say f(x)/x is not the same as 3x+7? I feel like I don't understand exactly why, since I don't feel like we've really divided by zero.

Answer 4

\[\large g(x) = \dfrac{ f(x)}{x} = \begin{cases}
3x+ 7 & x\neq 0 \\
? & x = 0 \\

\end{cases}
\]

Answer 5

let f(x)=3x^2+7x
and g(x)=x

f(x) / g(x) is a new function and its different than the inear 3x+7
sketch both to see that

Answer 6

For instance, what if I had been given \[\Large f(x)=3x^2+7x\] Is this the same thing as \[\Large f(x)=x*(3x+7)\] or I unable to plug in x=0 to this second equation? After all, what is (3x+7) really?

Answer 7

mmm idk why wolfram sketch both to be the same -.-

Answer 8

I think a smiliar question is : whats the value of \(\large e^{ln x} \) when \(x = -2\) ?

Answer 9

f(x)/x = 3x+7
and at x=0, it is 7

Answer 10

what the doubt ?

Answer 11

hartnn it has a removable discontinuity at x = 0 right

Answer 12

Yes exactly. How much of this is "true" to the math and how much of this is just arbitrary problems that come up because of a faulty system of keeping track of our equations?

For example, I don't think anyone would argue with me here about this statement.
\[\large \sin(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}\]

However dividing both sides by x and evaluating them at x=0 gives two different answers. \[\large \frac{ \sin(x)}{x}=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)!}\] The left one is not able to be evaluated and the right one is 1. Which is the value of the limit.

But I just feel like limits are a way of simply "fixing" a broken notation, not really doing anything of any actual interest.

Answer 13

it?
you mean f(x)/x ?
it does not habe discontinuity at all

Answer 14

g(x) = f(x)/x = 3x+7
g(x) is continuous

Answer 15

@hartnn Yes, that's what I believe. Similarly if we had said f(x)=sin(x) and to evaluate f(x)/x I believe that this too is a continuous function everywhere. I think our ability to divide sin(x) by x is simply not easy to account for with our notation because we would have to invent a new function.

Answer 16

there was a zero at x=0 for f(x), it was removed by dividing by x

Answer 17

thats right kainui

Answer 18

I want to believe you but I remember when I took calculus 1 finding the limit as x approaches 0 of sin(x)/x. In my mind this is a completely foolish task because this point is really not any more important than any other point. It just happens to be a problem with the notation.

Answer 19

f(x)/x is EXACTLY same as 3x+7

Answer 20

\(\large \sin x \ne x\)

Answer 21

\(\Large \sin x = x, x\to 0\)

Answer 22

\[\large \frac{ \sin(x)}{x}=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)!}\]

if this were true, then we get

\(\sin x = x \) which is just nonsense

Answer 23

f(x)/x is a linear continuous function defined by 3x+7

Answer 24

@ganeshie8 that's the taylor series of sin(x)... It's definitely true.

http://www.wolframalpha.com/input/?i=taylor+series+sin+x

Answer 25

you can't evaluate
\(\large \frac{ \sin(x)}{x}=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)!}\)
at x=0, you will get an indeterminate form 0/0
hence we take the limit and say that limiting value = 1

Answer 26

I agree with you @hartnn but I don't feel like that your answer is justified and I dont' exactly know how to justify it other than it's my gut instinct.

Answer 27

Wait, the first term of that series is 1 and the rest depends on x. \[\Large \frac{\sin x}{x}=1+\sum_1^\infty \frac{(-1)^nx^{2n}}{(2n+1)!}\] So this means \[\Large \frac{\sin 0}{0}=1+\sum_1^\infty \frac{(-1)^n0^{2n}}{(2n+1)!}=1\]

Answer 28

add the limit
lim x->0 sin x/x = 1+0 =1

Answer 29

Yeah, that's what I'm debating. The whole limit part seems fake to me.

Answer 30

If f(x)=3x^2+7x then what's f(x)/x=3x+7 which is linear continuous function, any doubts here ?

sin x/x is not defined at x=0, so we take limit x->0 of sin x/x and we prove its value = 1

Answer 31

Alright so by your logic pretend that this power series doesn't represent sin(x). It is simply a polynomial, which is what it is.

\[\Large f(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}\] Ok now evaluate f(x)/x at x=0

We don't have to take a limit just like you did earlier with f(x)=3x^2+7x either.

Answer 32

`If f(x)=3x^2+7x then what's f(x)/x=3x+7 which is linear continuous function, any doubts here ?`

f(x)/x = 3x+7 everywhere except at x = 0, we don't know what f(x)/x equals at x = 0 - its not defined yet.

Answer 33

right, f(x)/x does equal 1, without the need of taking any limit

the need of taking limits comes because when we use the sine function for f(x) and divide it by x, at x=0, we get a 0/0 and THAT requires the need of limits

Answer 34

http://www.wolframalpha.com/input/?i=domain+%283x%5E2%2B7x%29%2Fx

Answer 35

If I rewrite it, then perhaps it will be clearer.\[\Large f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\]

Now earlier you said:

You said
"f(x)/x = 3x+7
and at x=0, it is 7"

How can we not do this? I'm sort of confused now as to where you stand.

Answer 36

Ahh ok I think we're on the same page now.

Answer 37

f(x)/x = 3x+7 everywhere except at x = 0 <<<NOT actually

3x^2 +7x /x is defined every where EXCEPT x=0

\(\Large \dfrac{3x^2 +7x}{x} \ne 3x+7 \)
^^ they both are NOT the same function!

Answer 38

Ok, this is my real question. Why? I just don't really see the reason we should leave a hole behind when it naturally fills itself in.

Answer 39

infact its illegal to say that f(x)/x = 3x+7 ....we need to define f(x)/x at x=0 , as ganeshie said.

Answer 40

It seems like a completely made-up problem to me. Is there a physical reason why we should consider this? Even if you were to integrate over the function with or without the hole taken out, you would still get the same answer.

Answer 41

Kainui, I see you want to remove the concept of removable discontinuties from Math.

Answer 42

Oh philosophy... it will never be put to rest ^_^

Despite 'not feeling like we really divided by zero' it all boils down to the fact that x is unknown...

Answer 43

I want to remove the concept of limits from math actually.

Answer 44

f(x) = 3x^2 +7x has a ZERO at x=0

f(x)/x = (3x^2 +7x)/x has a pole and a zero at x=0

3x+7 has no pole or zero at x=0

Answer 45

We need to ask ourselves WHY
\[\Large \frac{3x^2+7x}{x}\]
is apparently equal to
\[\Large 3x + 7\]

Answer 46

Yeah, good question.

Answer 47

Humour me guys, I know this is basic stuff, but... first, we factor out the common factor x
\[\Large \frac{x(3x+7)}{x}\]

Answer 48

And then cancel it out...

But then again... factoring out only makes sense if that which you're factoring out is not zero... right?

Answer 49

BINGO !

Answer 50

I don't know, where does this concept come from in the first place? Isn't this really the origin of the problem? This just feels circular to me.

Answer 51

Okay, so at least we're agreed that we've reduced the question to "why we cannot factor out a zero" yes? ^_^

Answer 52

\(\Large \frac{x(3x+7)}{x}, x\ne 0 \\ \Large =3x+7 , x\ne 0\)

Answer 53

Well I feel to some extent, we might be saying that

\[\large 3x^2+7x \ne x*(3x+7)\]
but I'm not quite sure.

Yeah sure continue @terenzreignz

Answer 54

Much like how maths was set in motion... we need principles... we need a solid foundation...

We can all agree that you cannot divide by zero, yes?

Answer 55

right! because you are dividing by x, thats why x cannot be 0

Answer 56

Well YOU can't, but I can.
I'm that awesome >:)

lol
kidding ^_^

Answer 57

No, I think it depends on the context of the situation. All derivatives are in my opinion are simply dividing by zero within a certain context.

Answer 58

No they're not... they're limits as the divisor gets closer and closer to zero...

but I suppose that's open to interpretation ^_^

Answer 59

Yes, see this is what I'm talking about. To talk about anything else is to get around the subject of interest and will only reexplain things I already "know".

Answer 60

But differentiation isn't really straightforward dividing by zero, is it?

Otherwise, why the limits? ^_^

Answer 61

Oh wait, I have an idea....

We can all agree that you can't divide a NONZERO by zero then?

So that rules out differentiation, which is essentially zero/zero...

Answer 62

I think I can probably agree with that.

Answer 63

Similarly f(0)=0 and x=0 so our problem was also a case of 0/0.

Answer 64

Still wondering where to go with all this XD
Let me have a good think...

Answer 65

Never mind... the problem really IS with 0/0.

Answer 66

I feel like the whole thing is entirely a problem with our notation and limits are simply a way of doing mental gymnastics to convince ourselves of things. For instance, what's the slope of a vertical line? You can't express this with y=mx+b? Why is this? Should we take a limit here too? I feel like it's not a "real" concept since we can simply draw the vertical line.

Answer 67

Maybe the problem with zero/zero is that it can be any number?

Answer 68

Well that's what I'm saying. 0/0 has to be given a context, otherwise it's nonsense. It's like saying x can be any number. You have to use the concept of a variable to create equations and things, it doesn't really exist on its own.

Answer 69

Perhaps the rules we set for maths forbid the definition of 0/0.

Answer 70

isnt 0/0 just equal to 1?

Answer 71

No. 0/0 is an indeterminant form that has no definition.

Answer 72

So 0/0 is indeterminate (it could be any value)

Answer 73

Exactly. Just like x can be any value.

Answer 74

Okay... what about this:
Suppose 0/0 has some value.

\[\large \frac00 = k \qquad\qquad\qquad k \in \mathbb{R}\]

Answer 75

Ok.

Answer 76

Then let's multiply 1/0 on both sides.
\[\large \frac00 \cdot \frac10 = \frac k 0\]

Answer 77

I'm not saying 1/0 can have a value, only 0/0 can.

Answer 78

granted... sorry... rethinking XD

Answer 79

Haha this is fun, thanks for taking the time to help me out.

I think a physical problem where dividing by something that's zero and showing how leaving it in the domain on accident causes us to make a false assumption would be best since I believe math comes from physical observation.

Answer 80

Let me try with assuming \[\large \frac00 = k\]

Answer 81

well isnt f(x) = y?

Answer 82

First case, k is nonzero. Let's divide both sides by k.
\[\Large \frac{0}{0\cdot k}= 1\]

Answer 83

Everything still legit, right, Kai? :D

Answer 84

Seems alright so far. =) However it does seem sort of odd from my point of view since 0/0=k doesn't really define anything just like saying x=k and that k is an element of the reals doesn't really pin down much about what x is.

Answer 85

this question is such a troll

Answer 86

alright so far is good enough for me.

Now, by DEFINITION, \(\large 0\cdot k = 0\)
So we have:
\[\Large \frac 00 = 1\]
And we know 0/0 is k.
So necessarily, we have
\[\Large k = 1\]

Answer 87

from your explanation @terenzreignz k would equal any random number

Answer 88

That is a good observation, and that's the problem that plagues the enigmatic 0/0... it could be any number.

Answer 89

Only this time, BECAUSE it could be any number, it turns into a contradiction of the first assumption (namely, that k is nonzero)

Answer 90

This seems a little fishy to me since 0/0 is a number. Usually when I see something like this I am convinced, but this time I'm really not. Ok, maybe that's a personal problem, but still this just seems like a technicality and not really resolving it.

Answer 91

So far, I've only demonstrated that k cannot be nonzero... so what if it's zero?
Suppose \[\Large \frac00 = 0\]

Answer 92

In the context of sin(2x)/x evaluated at x=0 then we find that this version of 0/0 is really 2. Similarly if we look at sin(2x)/(2x) then at x=0 this version of 0/0 is really 1. This is really just saying 0/0=2 and 0/0=1 for the first and second cases, and it still seems to work out in each respective context.

Answer 93

hmmm

Answer 94

I guess this is why I don't really feel convinced. I mean, I want to believe, I just can't. I'm trying to sort of see what a limit really is and why it is necessary. It feels made up, I'm sorry that I can't be more mathematical than this; it just simply feels like there is a problem with notation and I don't think wording it within the broken system itself is really able to fix it.

Or maybe I just need to let it go and forget about it. Bleh.

Answer 95

We can't simply evaluate sin(2x)/x at x = 0 right? Only find its behaviour at points in its immediate vicinity? :/

Answer 96

Oh, never mind then XD

Answer 97

ohhh Einstein please help us... ㅜ.ㅜ