Can one construct a function having the following properties ?

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A function f(x): \mathbb{R} \to \mathbb{R} can be constructed such that \lim_{x \to 0} x f(x) = a, where a is a non-zero constant. The discussion highlights that substituting limits with functions can lead to a valid equation, allowing the definition of f(x) that meets the limit condition. An example provided is f(x) = 1/x for x ≠ 0 and f(0) = 0, which satisfies the limit requirement. It is noted that such a function cannot be continuous at x = 0 if the limit is non-zero. The exploration of ε(x) and g(x) functions is suggested to further refine the construction of f(x).
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Is there a function f(x): \mathbb{R} \to \mathbb{R} such that \lim_{x \to 0} x f(x) = a \neq 0.
 
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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?
 
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.
 
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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.
 
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.
 
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.
 

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