MHB Limit Calculation Help: x^3(ln(x))^2

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To solve the limit \(\lim_{x\to 0^{+}}x^{3}\cdot (\ln(x))^{2}\), it is suggested to rewrite it as \(\lim_{x\to0^{+}}\frac{\ln^2(x)}{x^{-3}}\) to transform the indeterminate form from \(0\cdot\infty\) to \(\frac{\infty}{\infty}\). This allows for the application of L'Hôpital's rule. By differentiating the numerator and denominator, the limit can be evaluated more easily. Further assistance is requested to proceed with the calculations.
Yankel
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Hello

I need some help with this limit. I know I should use the L'hopital rule, but not sure how to do it...any help will be appreciated !

\lim_{x\to 0^{+}}x^{3}\cdot (ln(x))^{2}
 
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I notice this is the indeterminate form $0\cdot\infty$, so I would recommend writing it as:

$\displaystyle \lim_{x\to0^{+}}\frac{\ln^2(x)}{x^{-3}}$

Now we have the indeterminate form $\dfrac{\infty}{\infty}$ and we may apply L'Hôpital's rule.

Can you proceed?
 
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