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I’m trying to give at least some partial answers for one of my own questions (this one).

There the following arose:

$\hskip1.7in$ Why is $\lim_{x \to 0} {\rm li}(n^x)-{\rm li}(2^x)=\log\left(\frac{\log(n)}{\log(2)}\right)$?

Expanding at $x=0$ doesn’t look reasonable to me since ${\rm li}(1)=-\infty$

and Wolfram only helps for concrete numbers, see here for example.

Would a “$\infty-\infty$” version of L’Hospital work?

Any help appreciated.

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Thanks,

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- Evaluate $\lim_{x \to \infty} \frac{(\frac x n)^x e^{-x}}{(x-2)!}$
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$$

\begin{align}

\lim_{x\to0}\int_{2^x}^{n^x}\frac{\mathrm{d}t}{\log(t)}

&=\int_{x\log(2)}^{x\log(n)}\frac{e^u}{u}\mathrm{d}u\\

&=\lim_{x\to0}\left(\color{#C00000}{\int_{x\log(2)}^{x\log(n)}\frac{e^u-1}{u}\mathrm{d}u}

+\color{#00A000}{\int_{x\log(2)}^{x\log(n)}\frac{1}{u}\mathrm{d}u}\right)\\

&=\color{#C00000}{0}+\lim_{x\to0}\big(\color{#00A000}{\log(x\log(n))-\log(x\log(2))}\big)\\

&=\log\left(\frac{\log(n)}{\log(2)}\right)

\end{align}

$$

**Note Added:** since $\lim\limits_{u\to0}\dfrac{e^u-1}{u}=1$, $\dfrac{e^u-1}{u}$ is bounded near $0$, therefore, its integral over an interval whose length tends to $0$, tends to $0$.

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