# Can one construct a function having the following properties ?

1. Apr 1, 2013

### funcalys

Is there a function $f(x): \mathbb{R} \to \mathbb{R}$ such that $\lim_{x \to 0} x f(x) = a \neq 0$.

Last edited by a moderator: Apr 1, 2013
2. Apr 1, 2013

### micromass

Staff Emeritus
Such a function exists. What did you try already??

What can you tell about $\lim_{x\rightarrow 0} f(x)$. Is it possible that this limit is finite?

3. Apr 1, 2013

### CompuChip

If f(x) = 1/x then x f(x) = 1, except that doesn't work in x = 0. But maybe that will get you started.

Last edited: Apr 1, 2013
4. Apr 1, 2013

### CompuChip

Haha this thread is getting confusing.

In the end the answer is: yes, there is a function $f(x): \mathbb{R} \to \mathbb{R}$ such that $\lim_{x \to 0} x f(x) = a \neq 0$.

5. Apr 4, 2013

### Tosh5457

Yes, the trick with limits is to substitute them with functions so you get a normal equation:

Write xf(x) + ε(x) = g(x)

with ε(x) and g(x) such as
lim (x->0) ε(x) = 0 and lim (x->0) g(x) = a

so that lim (x->0) ( xf(x) + ε(x) = g(x) ) <=> lim (x->0) xf(x) = a

Now define a ε(x) and g(x) that satisfy those conditions and you can find a function f(x) that satisfies that limit.

If you want x = 0 to be in the domain of f(x), you should also set the conditions ε(0) = 0 and g(0) = 0.

6. Apr 4, 2013

### HallsofIvy

Staff Emeritus
The function f(x)= 1/x if $x\ne 0$, f(0)= 0 is a perfectly good function that maps all R, one to one, onto R, such that $$\lim_{x\to 0} xf(x)= 1$$. Of course, a function such that $$\lim_{x\to 0} xf(x)$$ is non-zero cannot be continuous at x= 0.