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.
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