How to Prove 2x is Continuous

dcl

How do I show that f(x)=x^2 is continuous at any given point, say x=3.

Thank you.

arildno

You show that the function's limit at (x=3) equals the value of the function at x=3 (that would be 9)

dcl

thanks...
Using the left and right limits at x=3 yeh?
My notes have another method involving a delta-epsilon argument, how would that work?

franznietzsche

Actually the easiest way to prove continuity at all values is to show that the derivative is always defined, differentiability always implies continuity (note the converse is not always true.). So for f(x) = x^2, you get f'(x) = 2x, which is defined for all values of x thus f(x) is continuous across the interval (-infinity,infinity)

arildno

All right:
Assume [tex]|x-x_{0}|\leq\delta[/tex]

We must show that the function value at [tex]x_{0}[/tex] equals the limit value.
That means, we must estimate the expression [tex]|x^{2}-x_{0}^{2}|[/tex] :
[tex]|x^{2}-x_{0}^{2}|=|x-x_{0}||x+x_{0}|\leq\delta|x+x_{0}|[/tex]

Note that at this moment, I haven't bothered to introduce "epsilon" explicitly.
What's important at this stage is to find a bound in terms of delta.

We now have:
[tex]|x+x_{0}|\leq|x|+|x_{0}|,|x-x_{0}|\leq\delta\to|x|\leq|x_{0}|+\delta[/tex]

Hence, we have:
[tex]|x^{2}-x_{0}^{2}|\leq\delta^{2}+2\delta|{x}_{0}|[/tex]

Now, clearly if both [tex]\delta^{2},2\delta|{x}_{0}|\leq\frac{\epsilon}{2}[/tex],
then we have:
[tex]|x^{2}-x_{0}^{2}|\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon[/tex]

Hence, we should set [tex]\delta=min(\sqrt{\frac{\epsilon}{2}},\frac{\epsilon}{4|x_{0}|})[/tex]

dcl

I'm sorry, I don't really understand the 'method' or the reasoning behind that. I need to learn it, but can't for the life of me understand it at the moment.

Where did the epsilon in the 3rd last line come from?

HallsofIvy

I'm sorry, I don't really understand the 'method' or the reasoning behind that. I need to learn it, but can't for the life of me understand it at the moment.
Where did the epsilon in the 3rd last line come from?

It came from the definition of limit. If you are trying to understand "continuity" I strongly recommend that you go back and understand "limit" first.

matt grime

Actually the easiest way to prove continuity at all values is to show that the derivative is always defined, differentiability always implies continuity (note the converse is not always true.). So for f(x) = x^2, you get f'(x) = 2x, which is defined for all values of x thus f(x) is continuous across the interval (-infinity,infinity)

*cough* where have you proved that the derivative exists and is equal to 2x? You are assuming x^2 is diffble, and in that case you might as well assume it is continuous, mightn't you?

arildno

I'm sorry, I don't really understand the 'method' or the reasoning behind that. I need to learn it, but can't for the life of me understand it at the moment.
Where did the epsilon in the 3rd last line come from?

Let's study the limit concept in some detail:

We say that a number L is the limit of a function f at a point [tex]x_{0}[/tex]
if the following property holds:

Given any [tex]\epsilon>0[/tex] there exists a number [tex]\delta>0[/tex] so that for every [tex]x\neq{x}_{0}[/tex] which satisfy [tex]|x-x_{0}|<\delta[/tex] it follows that [tex]|f(x)-L|<\epsilon[/tex]

L is symbolized by the expression [tex]\lim_{x\to{x}_{0}}f(x)[/tex]

Comments:
1."We say that a number L is the limit of a function f ..if.."
Note that at no point are we given a method to derive L, but neither are we required to derive this number.
Rather, what we are given is a method to analyze whether or not an arbitrarily chosen number L can be said to be the limit of f (at the indicated point)

How or why you choose a particular L to analyze, is completely irrelevant!

2. "Given any [tex]\epsilon>0[/tex] "
This means that we must show the result for EVERY [tex]\epsilon>0[/tex]
In order to so, we need to formulate a bound for the expression [tex]|f(x)-L|[/tex] which can be related to an arbitrary [tex]\epsilon[/tex]
Otherwise, we will be unable to derive the desired result!

3. "there exists a number [tex]\delta>0[/tex]"
At no place is it stated that only one such number exists, what we're after is the existence of at least one such number.

4." so that for every [tex]x\neq{x}_{0}[/tex]"
This is a rather technical issue which is vital to the distinction between the limit value of a function at a given point and the function value at the same point.
For example, it is needed in order to have existence of limits at points where the function is discontinuous, or at points where the function f is undefined (i.e. has no specified function value).
As an example, we want the function f(x)=0 for x not 0, f(0)=1 to have the limit at x=0 [tex]\lim_{x\to0}f(x)=0[/tex]

5. "there exists a number [tex]\delta>0[/tex] so that for every [tex]x\neq{x}_{0}[/tex] which satisfy [tex]|x-x_{0}|<\delta[/tex] it follows that [tex]|f(x)-L|<\epsilon[/tex]"

Here, we have been given the tool to analyze with ([tex]|x-x_{0}|<\delta[/tex] ) and the result we are seeking fulfilled: ([tex]|f(x)-L|<\epsilon[/tex])

What we need to do therefore, is to obtain so much information as possible about [tex]|f(x)-L|[/tex] with our aid [tex]|x-x_{0}|<\delta[/tex]

In particular, we should first try to gain a bound on the form:
[tex]|f(x)-L|<G(\delta)[/tex]
(Note that I express this bound in terms of [tex]\delta[/tex] and not [tex]\epsilon[/tex]!!)

The second phase is now to to find the bounds of [tex]\delta[/tex], so that
[tex]G(\delta)<\epsilon[/tex]
Then we are finished!

Note in particular that the bounds on [tex]\delta[/tex] will in general be functions of [tex]\epsilon[/tex] and [tex]x_{0}[/tex]

Hope this helped

dcl

Thanks for that. I think I've finally got some 'sense' on the matter.

e(ho0n3

Looking at this thread maid me realized that I have completely forgotten how the limits are computed. It appears to me that the definition of the limit does not enable you to determine what the value of the limit is. For example, if I made a program to calculate the limit of some function using the definition, I would only be able to approximate what the limit is. In other words, the only thing I can do is bound the value of the limit. This makes me wonder how programs like Mathematica and Maple calculate these things.

e(ho0n3

"There are no stupid questions, just stupid people."

Hurkyl

You can do some pretty magical things with algebra; the trick is just to find a way of encoding concepts like limits in a way that can be manipulated algebraically.

dcl

How would this work in a multivariable situation?
say for:
1/xy at (1,1)

arildno

The first thing to do in the multivariable case, is to make a proper generalization of the [tex]\epsilon,\delta[/tex]-formalism!

I will do so in the case of a scalar function f from [tex]\Re^{2}[/tex] ,the real plane, into the real line, [tex]\Re[/tex]
(That's what your example is about)

We say that a number L is the limit of f at a point [tex](x_{0},y_{0})[/tex] if for any [tex]\epsilon>0[/tex] there exists a number [tex]\delta>0[/tex], such that for any [tex](x,y)\neq(x_{0},y_{0})[/tex] fulfilling [tex]||(x,y)-(x_{0},y_{0})||<\delta[/tex] it follows that [tex]|f(x,y)-L|<\epsilon[/tex]

Note that since the domain is now 2-dimensional, the [tex]\delta[/tex]-"interval" is now a disk centered on [tex](x_{0},y_{0})[/tex]

It is a good exercise to use this definition and try to establish the continuity of the given f by yourself first..

franznietzsche

*cough* where have you proved that the derivative exists and is equal to 2x? You are assuming x^2 is diffble, and in that case you might as well assume it is continuous, mightn't you?

umm...use the definition of the derivative as the ratio of [tex] \frac{\Delta f}{\Delta x} [/tex] and you will get that answer, always. So its not an assumption. Try the theorems of differential calculus.

master_coda

umm...use the definition of the derivative as the ratio of [tex] \frac{\Delta f}{\Delta x} [/tex] and you will get that answer, always. So its not an assumption. Try the theorems of differential calculus.

But you never provided a proof that the derivative was defined. The reasoning you provided treated the fact that f'(x)=2x as an obvious fact, not as something you have to prove.

Plus, the derivative is not defined as the ratio [itex]\frac{\Delta f}{\Delta x}[/itex].

matt grime

umm...use the definition of the derivative as the ratio of [tex] \frac{\Delta f}{\Delta x} [/tex] and you will get that answer, always. So its not an assumption. Try the theorems of differential calculus.

So in order to prove it is continuous you are requiring that we prove from first principles that is differentiable. And that is simpler is it than proving that it is continuous directly? Seems about the same if not more complicated to me.

mathwonk · Jul 29, 2004

this seems pretty well covered but I am not deterred from trying to make it look easier.

i.e. why is x^2 continuous?

Recall a function f is continuous at x, if when z approaches x, then f(z) approaches f(x).

So let z = x+h. then z approaches x if h approaches zero.

Now just work out z^2 and compare it to x^2.

I.e. z^2 = (x+h)^2 = x^2 + 2xh + h^2.

Now what happens as h approaches zero? Well obviously (since x is fixed) then also 2xh approaches zero and so does h^2. So z^2 approaches x^2, i.e. f(z) does approach f(x) and you are done.

You do this when x = 3 and see what happens. E.g. how small does h have to be in that case for z^2 to be closer to 3^2 than 1/10?

mathwonk

the multivariable case is exactly the same only harder.

Take f(x,y) = 1/(xy), and set z = x+h, and w = y+k, and let both h,k, approach zero, and compare f(z,w) with f(x,y).

We get f(z,w) = 1/(x+h)(y+k) and I have to compare this to 1/xy, but this is a pain because fractions are hard./

Anyway lets try it: 1/(x+h)(y+k) - 1/(xy) = (common denominator)

= [xy - (x+h)(y+k)]/[(xy)(x+h(y+k)] = -(hy+kx+hk)/[(xy)(x+h(y+k)].

Now just look at this mess, and let both h,k approach zero. The top approaches zero and the bottom approaches (xy)^2, so as long as neither x nor y is zero, this approaches zero.

I.e. we showed that f(z,w) - f(x,y) approaches zero as (z,w) approaches (x,y), so in other words f(z,w) approaches f(x,y) as (z,w) approaches (x,y), (when both x and y are non zero).

This is harder but can you see how small h and k should be to make f(z,w)-f(x,y) less than 1/10? You do not need to find the best possible answer but you can overdo it. I.e. just find a real small value that will guarantee it works. I.e. I am taking epsilon equal to 1/10 and asking you for some corresponding delta.

Monty Dersecting

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