A function that equals its limit (?)

**rainer** · January 11th 2010, 05:39 AM

Show me a function f(x) such that $f(x)=\lim_{x\to0}f(x)$

Actually this really more of a question I have than a challenge. I do have a solution but am challenging inquisitively. If that gets on everyone's nerves just kindly ignore this and I will post it later on as a challenging inquisition rather than an inquisitive challenge.

**Chris L T521** · January 11th 2010, 10:31 AM

Originally Posted by rainer

Show me a function f(x) such that $f(x)=\lim_{x\to0}f(x)$

Actually this really more of a question I have than a challenge. I do have a solution but am challenging inquisitively. If that gets on everyone's nerves just kindly ignore this and I will post it later on as a challenging inquisition rather than an inquisitive challenge.

The obvious case is when $f(x)=c$ for $c\in\mathbb{R}$ . I'm not sure what else to suggest, since $\lim_{x\to 0}f(x)$ should be a numerical value...

**Drexel28** · January 11th 2010, 01:18 PM

Originally Posted by rainer

Show me a function f(x) such that $f(x)=\lim_{x\to0}f(x)$

Actually this really more of a question I have than a challenge. I do have a solution but am challenging inquisitively. If that gets on everyone's nerves just kindly ignore this and I will post it later on as a challenging inquisition rather than an inquisitive challenge.

$f(x)=0$ .

As Chris said though, in general this is not true. In, fact the question tells you the answer. If $f(x)=\lim_{x\to x_0}f(x)$ for all $x_0\in\text{Dom }f$ then $f(x)=f\left(x_0\right)$ (assuming this function is continuous) and so it's constant.

**rainer** · January 12th 2010, 05:40 AM

Originally Posted by Drexel28

$f(x)=0$ .

As Chris said though, in general this is not true. In, fact the question tells you the answer. If $f(x)=\lim_{x\to x_0}f(x)$ for all $x_0\in\text{Dom }f$ then $f(x)=f\left(x_0\right)$ (assuming this function is continuous) and so it's constant.

(What is "Dom f?")

Right so this is exactly my question. f(x) has to be a numerical value and a function for $f(x)=\lim_{x\to x_0}f(x)$ to work.

And yet a numerical value AND a function is exactly what we get as the roots of some second degree polynomials.

**Drexel28** · January 12th 2010, 07:26 AM

Originally Posted by rainer

(What is "Dom f?")

Right so this is exactly my question. f(x) has to be a numerical value and a function for $f(x)=\lim_{x\to x_0}f(x)$ to work.

And yet a numerical value AND a function is exactly what we get as the roots of some second degree polynomials.

$\text{Dom }f$ is the domain. I don't see what the problem is. Suppose we have thta $f(x)=\lim_{x\to0}f(x)$ . Regardless of whether or not the function is continuous, and it must be, we have that $f(x)=c$ for some $c$ (specifically the value of the limit) if the limit exists. I don't get the reference to quadratics

**rainer** · January 13th 2010, 06:08 AM

$f(x)\equiv y$

The roots of y in the equation $y^2x^2+y(2x^2-1)+x^2-1=0$ (which is the equation plotted in the attachment)
are $-1$ and $\frac{1}{x^2}-1$

So, $\lim_{x\to\infty} y$ i.e. $\lim_{x\to \infty}\frac{1}{x^2}-1=-1$

(The limit was as x goes to infinity, not 0. REALLY SORRY about that.)

and yet the other root of y euals this limit, equals -1. So, can't it be said that the function y equals its limit?

**Drexel28** · January 13th 2010, 07:40 AM

Originally Posted by rainer

$f(x)\equiv y$

The roots of y in the equation $y^2x^2+y(2x^2-1)+x^2-1=0$ (which is the equation plotted in the attachment)
are $-1$ and $\frac{1}{x^2}-1$

So, $\lim_{x\to\infty} y$ i.e. $\lim_{x\to \infty}\frac{1}{x^2}-1=-1$

(The limit was as x goes to infinity, not 0. REALLY SORRY about that.)

and yet the other root of y euals this limit, equals -1. So, can't it be said that the function y equals its limit?

When you say $f(x)=\lim_{x\to x_0}f(x)$ that literally means " $f(x)$ is a function that takes the value of $\lim_{x\to x_0}f(x)$ at every point of its domain."

**rainer** · January 14th 2010, 05:53 AM

Originally Posted by Drexel28

When you say $f(x)=\lim_{x\to x_0}f(x)$ that literally means " $f(x)$ is a function that takes the value of $\lim_{x\to x_0}f(x)$ at every point of its domain."

Thanks. And yet this is true, if it is true that $f(x)=-1$ AND $f(x)=\frac{1}{x^2}-1$

And yet it is untrue, as you point out, for the very same reason. In other words this is a logical contradiction. And a contradiction on two counts: 1) because one root is a constant, and the other is a function, and yet both roots must be defined at the outset as EITHER a function OR a constant, not both; and 2) because the domain of one root is undefined or mathematically meaningless vis-a-vis the domain of the other root (because it involves 1/infinity).

How does one resolve the contradiction? How do you qualify "AND" to resolve it?

**HallsofIvy** · January 14th 2010, 06:59 AM

Any continuous function has the property that $\lim_{x\to x_0}f(x)= f(x_0)$ , that's the definition of "continuous". It is true for all $x_0$ for any function, f, that is continuous for all real numbers.

That is NOT true for $f(x)= \frac{1}{x^2}- 1$ because that function is not defined, and so not continuous, at x= 0.

I have no idea what you are trying to say.

**rainer** · January 15th 2010, 04:30 AM

Originally Posted by HallsofIvy

I have no idea what you are trying to say.

Surely, the problem is rather that I am not grasping the full import of what you are all saying. So, thanks for your patience and comments. I will ruminate on the input I've received in this thread and will no doubt eventually come around to a proper understanding (both of what you're saying and of what I'm trying to get at).

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