# Thread: A function that equals its limit (?)

1. ## A function that equals its limit (?)

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.

2. 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...

3. 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.

4. 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.

5. 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

6. $f(x)\equiv y$

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?

7. 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."

8. 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?

9. 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.

10. 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).