# L'Hopital's rule of x*ln(x)-x

#### blessedcurse

1. The problem statement, all variables and given/known data
What is the value of xln(x)-x when x=0?

2. Relevant equations
I'm assuming you do L'Hopital's

3. The attempt at a solution
I'm assuming you factor out the x, leaving:

x(ln(x)-1)

but that's still not in the form of $$\frac{\infty}{\infty}$$ or $$\frac{0}{0}$$

Would you do:

$$lim_{x\rightarrow\infty}$$$$\frac{(ln(x)-1)}{x^{(-1)}}$$

=$$lim_{x\rightarrow\infty}$$$$\frac{(1/x)}{1}$$

=0

??

#### gabbagabbahey

Homework Helper
Gold Member
1. The problem statement, all variables and given/known data
What is the value of xln(x)-x when x=0?

2. Relevant equations
I'm assuming you do L'Hopital's

3. The attempt at a solution
I'm assuming you factor out the x, leaving:

x(ln(x)-1)

but that's still not in the form of $$\frac{\infty}{\infty}$$ or $$\frac{0}{0}$$

Would you do:

$$lim_{x\rightarrow\infty}$$$$\frac{(ln(x)-1)}{x^{(-1)}}$$

=$$lim_{x\rightarrow\infty}$$$$\frac{(1/x)}{1}$$

=0

??
$$\lim_{x\to 0} x\ln(x)-x=(\lim_{x\to 0} x\ln(x))-(\lim_{x\to 0} x)$$

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