Solving Limit x->0 of f(x)/x: A Hint

[ttttrigg3r]

1. Homework Statement
Given: f(x+y)=f(x) + f(y)+x²y+xy²
limit$_{x->0}$$\frac{f(x)}{x}$=1

Homework Equations

What is f(0)?

The Attempt at a Solution

The answer is f(0)=0 because as x->0 then f(x)->0. so f(0)=0

That is not my attempt. That is the solution. I am stuck from the beginning and don't know where to start. Can someone give me a hint as to how to start ?

 

[gb7nash]

1. Homework Statement
Given: f(x+y)=f(x) + f(y)+x²y+xy²
limit$_{x->0}$$\frac{f(x)}{x}$=1

Homework Equations

What is f(0)?

The Attempt at a Solution

The answer is f(0)=0 because as x->0 then f(x)->0. so f(0)=0

That is not my attempt. That is the solution. I am stuck from the beginning and don't know where to start. Can someone give me a hint as to how to start ?

I'm not sure why a limit existing implies the value of the function. Unless I'm missing something, you don't need the limit at all. Look at the first piece of information. Let x = y = 0, so:

f(0+0)=f(0) + f(0)+0²*0+0*0², so...

 

[ttttrigg3r]

I don't think that would work. what if x=2 and y=-2. and you still don't know what f(x) is and f(y) is.

 

[stallionx]

1. Homework Statement
Given: f(x+y)=f(x) + f(y)+x²y+xy²
limit$_{x->0}$$\frac{f(x)}{x}$=1

Homework Equations

What is f(0)?

The Attempt at a Solution

The answer is f(0)=0 because as x->0 then f(x)->0. so f(0)=0

That is not my attempt. That is the solution. I am stuck from the beginning and don't know where to start. Can someone give me a hint as to how to start ?

f(0)=0

can be seen clearly when x,y are set to zero

 

[ttttrigg3r]

ok well if 2 people say it is then I will believe that. Thanks.

 

[vela]

The original answer you were given is incorrect or, at best, incomplete.

The statement that the only way the limit $$\lim_{x \to 0} \frac{f(x)}{x} = 1$$can exist is if $$\lim_{x \to 0} f(x) = 0$$is correct. Why? Because if f(x) didn't approach 0 as x went to 0, the quantity f(x)/x would diverge in the same limit, so the original limit couldn't exist.

However, saying$$\lim_{x \to 0} f(x) = 0$$is not the same as saying f(0)=0. The limit only tells you what the function is doing near the origin. It says absolutely nothing about what the function actually does at the origin. The function could be discontinuous at x=0 or even undefined, and the limit could still exist.

Now if you can use the given information about f(x) to show it's continuous, then yes, it follows that f(0) = 0 from the limit. A complete solution based on the limit would have to show f(x) is continuous.

 

[ttttrigg3r]

So are you saying that the problem lacks the information to find out f(0) and that the answer arrived upon, 0, is only partially complete? Should I even consider the first equation for f(x+y) when trying to find f(0), because there are others saying f(0) means that x and y are also zero.

 

[uart]

So are you saying that the problem lacks the information to find out f(0) and that the answer arrived upon, 0, is only partially complete? Should I even consider the first equation for f(x+y) when trying to find f(0), because there are others saying f(0) means that x and y are also zero.

The point Vela makes is that the given limit implies that $f(x) \rightarrow 0$ as $x \rightarrow 0$, but this is not quite the same thing as saying f(0)=0.

For an example that illustrates the difference, say $f(x) = \frac{1-\cos(x)}{x}$. Then the limit of f(x) as x goes to zero is certainly zero, but f(0) is undefined.

The other method that people have tried to point out to you is to replace both x and y with zero in the given definition of f(x+y).

f(0+0) = f(0) + f(0) + 0 + 0

which gives : f(0) = 2 f(0)

Subtracting f(0) from both sided gives f(0) = 0.

 

[uart]

BTW. Just out of interest, you can use the given data to find the first principles derivative of f(x) and hence establish an explicit function for f. You get,

$$f(x) = x + \frac{1}{3}x^3$$

 

[ttttrigg3r]

Thank you uart I understand that explanation very well.

Side question. because the denominator is x and x->0 wouldn't that make the equation undefined? In this case how would I approach this problem since there is no factor for me to cancel and f(x) is not a factor with numbers or anything I can use to cancel out x. How would I apply the limit to both top and bottom of the f(x)/x as x->0?

 

[ttttrigg3r]

@ uart
I calculated f'(x) using the definition of derivative to be x²+1
integrating to get f(x) would that add a constant C to f(x) or can that be omited?

 

[uart]

@ uart
I calculated f'(x) using the definition of derivative to be x²+1
integrating to get f(x) would that add a constant C to f(x) or can that be omited?

Yes that's correct. I used the given f(x+y) expression to find lim (f(x+h) - f(x)) / h and do the first principle derivative.

You get f'(x) = 1 + x^2 and hence f(x) = x + 1/3 x^3 + C, however the requirement that lim [f(x)/x] = 0 implies that C=0.

 

[NascentOxygen]

Given: f(x+y)=f(x) + f(y)+x²y+xy²

I find this function intriguing. I've been mulling this over since it was first posted, and I'm no nearer to understanding what it is I'm looking at.

(I understand the OP's question, and a bit of playing around with algebra soon answers his question.) My question is more fundamental: what does this f(x+y) look like? How can I represent it graphically?

Normally, a function of two variables is written f(x,y). This one takes only one parameter, so I guess that means it is a function of only one variable, e.g., f(w), yet it involves both x and y. Are x and y both independent variables? Is so, why isn't it a function of two variables, then?

Is this something in 3 dimensions, e.g., such that on the z axis we plot f(x+y)?

Maybe it's a recursive function, and y = f(x+y) ??

I'm hoping someone can clue me in on what's going on here.

 

[D H]

@Nascent: Think of g(x,y) = sin(x+y). There's nothing confusing about that, is there?

Or think of the definition of the derivative, the limit of (f(x+h)-f(x))/h as h tends to zero. There's nothing special here about that variable h. The limit of (f(x+y)-f(x))/y as y tends to zero is also f'(x). Once again we have a function of one variable, f(x) but we are evaluating it at the point x+h (or at the point x+y).

 

[NascentOxygen]

@Nascent: Think of g(x,y) = sin(x+y). There's nothing confusing about that, is there?

You haven't specified what y is.

I'm perfectly happy with y = g(x,w) = sin(x+w)
where x and w are independent variables.

If the normal convention applies, i.e., y=g(arg1,arg2) then I remain puzzled, because it's
y = g(x,y) = sin(x+y)

So, what precisely is y?

 

[D H]

What is x? It isn't specified, either. It is a variable. So is y.

 

[NascentOxygen]

Or think of the definition of the derivative, the limit of (f(x+h)-f(x))/h as h tends to zero. There's nothing special here about that variable h.

On the face of it, I'd expect that h would need to have the same units as x.

The limit of (f(x+y)-f(x))/y as y tends to zero is also f'(x).

I've never thought of it this way, and if y had the same units as x, this may be true. But if y is f(x), then in general it will not have the same dimensions as x, so a difficulty will be trying to add x + y. (x may be metres, and y metres³, for example, if the function is the volume of a sphere of radius x). Perhaps it isn't necessary to actually perform the addition, it might all come out in the wash. This will then hinge on being able to expand f(a+b)=...

 

[HallsofIvy]

You haven't specified what y is.

I'm perfectly happy with y = g(x,w) = sin(x+w)
where x and w are independent variables.

If the normal convention applies, i.e., y=g(arg1,arg2) then I remain puzzled, because it's
y = g(x,y) = sin(x+y)

So, what precisely is y?

y= sin(x+y) is not a definition, it is a "functional equation".