Evaluating $\lim_{x \to 1} \frac{1 - x^{1/\pi}}{1 - x^{1/e}}$ without de l'Hospital rule Is it possible to evaluate this limit without de l'Hopital's rule?
$$\lim_{x \to 1} \frac{1 - x^{1/\pi}}{1 - x^{1/e}}$$
 A: Using the fact that $ \lim\limits_{x\to 0}{\frac{\mathrm{e}^{x}-1}{x}}=1 $, we have : \begin{aligned}\Large \frac{1-x^{\frac{1}{\pi}}}{1-x^{\frac{1}{\mathrm{e}}}}=\frac{\exp{\left(\frac{\ln{x}}{\pi}\right)}-1}{\exp{\left(\frac{\ln{x}}{\mathrm{e}}\right)}-1}=\frac{\mathrm{e}}{\pi}\times\frac{\frac{\exp{\left(\frac{\ln{x}}{\pi}\right)}-1}{\frac{\ln{x}}{\pi}}}{\frac{\exp{\left(\frac{\ln{x}}{\mathrm{e}}\right)}-1}{\frac{\ln{x}}{\mathrm{e}}}}\underset{x\to 1}{\longrightarrow}\frac{\mathrm{e}}{\pi} \end{aligned}
Because $ \lim\limits_{x\to 1}{\frac{\exp{\left(\frac{\ln{x}}{\pi}\right)}-1}{\frac{\ln{x}}{\pi}}}=\lim\limits_{y\to 0}{\frac{\mathrm{e}^{y}-1}{y}}=1$, and $ \lim\limits_{x\to 1}{\frac{\exp{\left(\frac{\ln{x}}{\mathrm{e}}\right)}-1}{\frac{\ln{x}}{\mathrm{e}}}}=\lim\limits_{y\to 0}{\frac{\mathrm{e}^{y}-1}{y}}=1 $.
A: Why not to use
$$1-x^a=-a (x-1)+\frac{1}{2} \left(a-a^2\right) (x-1)^2+O\left((x-1)^3\right)$$
$$\frac {1-x^a}{1-x^b}=\frac{-a (x-1)+\frac{1}{2} \left(a-a^2\right) (x-1)^2+O\left((x-1)^3\right)}{-b (x-1)+\frac{1}{2} \left(b-b^2\right) (x-1)^2+O\left((x-1)^3\right)}$$
$$\frac {1-x^a}{1-x^b}=\frac{a}{b}+\frac{a  (a-b)}{2 b}(x-1)+O\left((x-1)^2\right)$$ which shows the limit and also how it is approached.
