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

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 [tex]\frac{\infty}{\infty}[/tex] or [tex]\frac{0}{0}[/tex]

Would you do:

[tex]lim_{x\rightarrow\infty}[/tex][tex]\frac{(ln(x)-1)}{x^{(-1)}}[/tex]

=[tex]lim_{x\rightarrow\infty}[/tex][tex]\frac{(1/x)}{1}[/tex]

=0

??
 

gabbagabbahey

Homework Helper
Gold Member
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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 [tex]\frac{\infty}{\infty}[/tex] or [tex]\frac{0}{0}[/tex]

Would you do:

[tex]lim_{x\rightarrow\infty}[/tex][tex]\frac{(ln(x)-1)}{x^{(-1)}}[/tex]

=[tex]lim_{x\rightarrow\infty}[/tex][tex]\frac{(1/x)}{1}[/tex]

=0

??
[tex]\lim_{x\to 0} x\ln(x)-x=(\lim_{x\to 0} x\ln(x))-(\lim_{x\to 0} x)[/tex] :wink:
 

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