Calculate the limit $\lim_{x\to 0}\frac{\arcsin(2x)}{\ln⁡(e-2x)-1}$ without using L'Hôpital's rule I need to calculate the limit without using L'Hôpital's rule:
$$\lim_{x\to 0}\frac{\arcsin(2x)}{\ln⁡(e-2x)-1}$$
I know that: $$\lim_{a\to 0}\frac{\arcsin a}{a}=1$$
But, how to apply this formula?

We have, $$\lim_{x \to 0} \frac{\arcsin 2x}{\ln(e-2x)-1} = \lim_{x \to 0}\frac{\arcsin 2x}{2x} \frac{2x}{\ln(e-2x)-1} = \lim_{x \to 0}\frac{\arcsin 2x}{2x}\frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}\times (-e)$$ This can be easily simplified to get the answer as $-e$. Hope it helps.
