What is the Limit of xln(x) - x as x Approaches 0?

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    L'hopital's rule
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SUMMARY

The limit of the expression xln(x) - x as x approaches 0 is evaluated using L'Hôpital's Rule. The discussion confirms that by factoring out x, the limit can be expressed as lim_{x→0} x(ln(x) - 1). The correct approach involves recognizing that as x approaches 0, ln(x) approaches -∞, leading to the limit being evaluated as 0. Therefore, the final conclusion is that lim_{x→0} xln(x) - x equals 0.

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Homework Statement


What is the value of xln(x)-x when x=0?

Homework Equations


I'm assuming you do L'Hopital's

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

??
 
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blessedcurse said:

Homework Statement


What is the value of xln(x)-x when x=0?

Homework Equations


I'm assuming you do L'Hopital's

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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