# 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.

• $$\lim_{x \to 0}\frac{\arcsin 2x}{2x}\frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}\times (-e)$$ I understand that $$\lim_{x \to 0}\frac{\arcsin 2x}{2x}=1$$, but why $$\frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}$$ is 1 too? – Dave Dec 26 '16 at 15:50
• – Dave Dec 26 '16 at 19:33

Hint:

$1=\ln e$

$\ln a-\ln b=\ln(a/b)$

$\lim_{h\to0}\frac{\ln(1+h)}h=1$

## Hint:

$$\frac{\arcsin(2x)}{\ln(e-2x)-1}=\frac{\arcsin(2x)}{2x}\frac{2x}{\ln(e-2x)-1}$$

Then combine with labbhattacharjee's answer for the logarithm.

The series expansion of $arcsin(2x)$ around $x=0$ is $$2x + 4x^3 +...$$

The series expansion of $ln(e-2x)$ around $x=0$ is $$1-\frac{2x}{e} - \frac{2x^2}{e^2} -...$$

As $x$ approaches $0$, all higher order terms vanish. Keeping the approximation up to first order:

$$\lim {x \to 0} \frac{arcsin(2x)}{ln(e-2x)-1} = \lim {x \to 0} \frac{2x}{-2x/e} = -e$$

Use equivalents:

$$\arcsin u\sim_0 u,\qquad \ln((1+u)\sim_0 u$$

and rewrite the function as $$\frac{\arcsin 2x}{\ln(e-2x)-1}=\frac{\arcsin 2x}{\ln\Bigl(1-\dfrac{2x}{\mathrm e}\Bigr)}.$$