Question

the limit of x^3lnx as x approaches 0 is indeterminate.evaluate it by using L'Hopital rule

Answer 1

so wats the question????????? not to sure what u're trying to find... it's x cubed what use the Equation thing

Answer 2

\[\lim_{x\to0^+}x^3\ln x=0\cdot(-\infty)\\
\lim_{x\to0^+}\frac{\ln x}{\frac{1}{x^3}}=\frac{-\infty}{\infty}\]
Now you can use L'Hopital's rule.

Answer 3

thank you. but then when i use the L'Hopital's rule i get non-stoping \[-\infty/\]

Answer 4

Applying L'Hopital's rule, you get
\[\begin{align*}\lim_{x\to0}\frac{\dfrac{1}{x}}{-\dfrac{3}{x^4}}&=\lim_{x\to0}\dfrac{1}{x}\left(-\dfrac{x^4}{3}\right)\\
&=-\frac{1}{3} \lim_{x\to0}x^3\\
\end{align*}\]

Answer 5

thanks , i get it now