Evaluate: $\lim_{x\to 1} \dfrac {m}{x^m -1} - \dfrac {p}{x^p-1}$

Question

Evaluate: $\lim_{x\to 1} \dfrac {m}{x^m -1} - \dfrac {p}{x^p-1}$

My Attempt: $$=\lim_{x\to 1} \dfrac {m}{x^m-1} - \dfrac {p}{x^p-1}$$ $$=\lim_{x\to 1} \dfrac {m(x^p-1)-p(x^m-1)}{(x^m-1)(x^p-1)}$$ $$=\lim_{x\to 1} \dfrac {m.x^p-m-p.x^m+p}{(x^m-1)(x^p-1)}$$

Answer 1

Note that we cannot use L'Hopital rule directly as the first step here, because we don't have a $\frac00$ or $\frac{\infty}{\infty} $ form yet.

But, simplifying: $$\lim_{x \to 1} \frac{m}{x^m-1} - \frac{p}{x^p-1}$$ we notice thatthe expression in $$= \lim_{x \to 1} - \frac{m-p+px^m - mx^p}{(x^m-1)(x^p-1)}$$ is of $\frac00$ form, so just use L'Hopital rule to get: $$\frac{p-m}2$$ as the answer.

Answer 2

Note that for $x\to 1 \implies \log x\to 0$

$$\dfrac {m}{x^m -1} - \dfrac {p}{x^p-1}=\frac{1}{\log x}\left(\dfrac {m \log x}{e^{m\log x -1}} - \dfrac {p \log x}{e^{p\log x}-1}\right)\to \frac{p-m}2 $$

indeed by Taylor's series

$$\left(\frac{e^y-1}{y}\right)^{-1}=\left( 1+\frac{y}2+o(y)\right)^{-1}=1-\frac{y}2+o(y)$$

$$\frac{1}{\log x}\left(\dfrac {m \log x}{e^{m\log x -1}} - \dfrac {p \log x}{e^{p\log x}-1}\right)=\frac{1}{\log x}\left( 1-\frac{m \log x}2-1+\frac{p \log x}2 +o(\log x)\right)=\\=\frac{p-m}2+o(1)\to \frac{p-m}2$$