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I have to compute $\lim \limits_{x\to 0}\frac{\ln(1+x^{2018} )-\ln^{2018} (1+x)}{x^{2019}} $. I tried to use L'Hopital's Rule, but it didn't work. I also tried to divide both the numerator and denominator by $x^{2018} $, but I still have a $\frac{0}{0}$ indeterminate form.

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  • $\begingroup$ I have edited the question, thanks $\endgroup$ – MathEnthusiast Mar 4 at 16:13
  • $\begingroup$ Have you also tried MacLaurin expansion of the logarithms at the numerator? (Second order should be enough.) $\endgroup$ – dfnu Mar 4 at 16:19
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By using the Maclaurin series of both terms the limit becomes $$\lim_{x \to 0} \frac{x^{2018}+o(x^{4036}) - (x^{2018}-1009x^{2019}+o(x^{2020}))}{x^{2019}}$$ $$=\lim_{x \to 0} \frac{1009x^{2019}+o(x^{2020})}{x^{2019}}$$ $$=\lim_{x \to 0}( 1009+o(x))$$ $$=1009$$

In fact, one can generalise this limit giving the following solution $$\lim_{x \to 0} \frac{\ln{(1+x^n)}-\ln^n{(1+x)}}{x^{n+1}}=\frac{n}{2}$$ for all $n\gt 1$.

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