1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $$ \lim_{x \to 0}\frac{\arcsin 2x}{2x}\frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}\times (-e)$$ 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? $\endgroup$ – Dave Dec 26 '16 at 15:50
  • $\begingroup$ Answer: math.stackexchange.com/questions/2072654/… $\endgroup$ – Dave Dec 26 '16 at 19:33
5
$\begingroup$

Hint:

$1=\ln e$

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

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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:

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

$\endgroup$
1
$\begingroup$

Use equivalents:

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.