Find Domin of f(x)= xlnx - LHopital Rule

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Hi

What is the domin of f(x)= xlnx.


I know that: domin of x is R and domin of lnx is x greater than 0 so the domin of them will be x greater than 0


but when I subtitude 0 on f(x) I find (o.\infty) and this form from LHopital Rule
so what shall I do I use the first tip or I use LHopital Rule to find the domin?
 
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ln x is defined only for positive real numbers, so the domain of your function is (0,inf). You can try and see of there is a limit of f as x tends to zero but this does not affect the domain.
 
If f(x) is defined on domain A and g(x) is defined on domain B, then f(x) times g(x) (or "plus" or "minus" g(x)) is defined on the intersection of sets A and B. In this case, the domain of ln(x), positive real numbers, is a subset of the domain of x, all real numbers. Their intersection is just the subset, all positive real numbers.

As Pere Callahan said, this has nothing at all to do with the limit or L'Hopital's rule.
 
You know, I really don't like this type of problem. Whenever you define a function you're required to specify its domain. So the question really ought to be asking for something like "maximal domain in R", because there is nothing wrong about defining the function f:{54, 97, 14654}->R by f(x)=xlnx.

I personally don't think that this kind of sloppiness is acceptable when you're just beginning to learn about functions, where this sloppiness is most present.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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