Hi, im trying to find the limit of xln(x) as it approaches 0+. I think the limit of ln(x) is undefined making it infinite right? But would the x at the begining make the limit approach 0?

Thank you.

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- Sep 23rd 2008, 07:42 PMcowboys111Quick limit help.
Hi, im trying to find the limit of xln(x) as it approaches 0+. I think the limit of ln(x) is undefined making it infinite right? But would the x at the begining make the limit approach 0?

Thank you. - Sep 23rd 2008, 07:48 PMChris L T521
Note that $\displaystyle \lim_{x\to{0^+}}x\ln x=\lim_{x\to{0^+}}\frac{\ln x}{\frac{1}{x}}$

you have the indeterminate case $\displaystyle \frac{\infty}{\infty}$.

We can now apply L'Hopital's rule:

$\displaystyle \therefore\lim_{x\to{0^+}}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to{0^+}}\frac{\frac{1}{x}} {-\frac{1}{x^2}}=\lim_{x\to{0^+}}-\frac{1}{\frac{1}{x}}=\lim_{x\to{0^+}}-x=\color{red}\boxed{0}$

Does this make sense?

--Chris - Sep 23rd 2008, 09:58 PMcowboys111
Yes it does, thank you.