Evaluating $\lim\limits_{x \to 0} \frac{\ln(1+x^{144})-\ln^{144}(1+x)}{x^{145}}$

Question

I've been trying to solve this limit using algebraic manipulations, L'Hospital's rule, approximations, but in vain. $$\lim\limits_{x \to 0} \frac{\ln(1+x^{144})-\ln^{144}(1+x)}{x^{145}}$$

Answer 1

A simple series expansion $$\log (1+z) = z - \frac{z^2}{2} + O(z^3)$$ followed by a binomial expansion will evaluate the limit quite easily.

Answer 2

Use the Taylor expansion for $ln(1+x)$: $x - \frac{x^2}{2} + \frac{x^3}{3} - ...$ and note that higher powers vanish more quickly as $x \rightarrow 0$.

$$\lim\limits_{x \to 0} \frac{x^{144}-O(x^{288})-(x^{144}-72x^{145}+O(x^{146}))}{x^{145}}$$

=

$$\lim\limits_{x \to 0} \frac{72x^{145}}{x^{145}}$$

= 72.

Answer 3

Be $n\in\mathbb{N}\setminus{1}$ .

$\displaystyle \frac{\ln(1+x^n)-(\ln(1+x))^n}{x^{n+1}}|_{x\to 0}=\frac{\ln(1+x^n)-x^n}{x^{n+1}}|_{x\to 0}+\frac{x^n-(\ln(1+x))^n}{x^{n+1}}|_{x\to 0}$

$\displaystyle =0+[\frac{z^n}{n!}](\frac{e^z-(1+x)^{z/x}}{x}|_{x\to 0})=[\frac{z^n}{n!}](e^z z)\frac{\frac{\ln(1+x)}{x}-\frac{1}{1+x}}{x}|_{x\to 0}=\frac{n}{2}$

Answer 4

$$\ln(1+x^{144})-\ln^{144}(1+x)=x^{144}-\frac{1}{2}x^{288}+o\left ( x^{288} \right )-\left ( x^{144}-72x^{145}+o(x^{145}) \right )$$ hence $$\lim\limits_{x \to 0} \frac{\ln(1+x^{144})-\ln^{144}(1+x)}{x^{145}}=72$$