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I've encountered this limit and I am stuck:

$$\lim \limits_{x\to0}{\frac{\frac{\ln(1+x)}{\arcsin{(x)}}-1}{x}}$$

My first thought were using some known limits, such as: $\lim \limits_{x\to0}{\ln(1+x)\over x}=1$ and $\lim\limits_{x\to0}{\arcsin{(x)}\over x}=1$ but no matter what I tried I got a limit of the form $0\over0$. So then I thought L'Hospital's rule might help, I used it to different forms of the limit but these seemed even more complicated, e.g.: $\lim \limits_{x\to0}{\frac{\sqrt{1-x^2}-1-x}{\sqrt{1-x^2}\arcsin{(x)}+x}}$.

I'd appreciate some hints on how to approach this limit. I am sure that it's possible to solve it just using these known limits and L'Hospital's rule. I just don't know how to manipulate the limit to a suitable form.

Thanks

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    $\begingroup$ just rewrite it with one numerator and one denominator and apply L'Hospital's rule twice. $\endgroup$ – Apurv May 24 '14 at 10:13
  • $\begingroup$ @Apurv I thought I might get away from taking the derivative of that, but it actually gives the result, thanks. $\endgroup$ – David May 24 '14 at 10:33
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So far, so good. Although, I notice that you've just omitted a factor that tends to 1 after using L'Hospital rule. But it's fine. You can continue by multiplying both numerator, and denominator by the numerator's conjugate, and arrive at:

$\begin{align*}\displaystyle \lim_{x \to 0} \frac{\sqrt{1 - x^2} - 1 - x}{\sqrt{1 - x^2}\arcsin x + x} &= \lim_{x \to 0} \frac{(1 - x^2) - (1 + x)^2}{(\sqrt{1 - x^2}\arcsin x + x) (\sqrt{1 - x^2} + 1 + x)} \\ &= \lim_{x \to 0} \frac{-2x - 2x^2}{(\sqrt{1 - x^2}\arcsin x + x) (\sqrt{1 - x^2} + 1 + x)} \\ &= \lim_{x \to 0} \frac{-2x (1 + x)}{(\sqrt{1 - x^2}\arcsin x + x) (\sqrt{1 - x^2} + 1 + x)} \\ &= \lim_{x \to 0} \frac{-2 (1 + x)}{\left(\frac{\sqrt{1 - x^2}\arcsin x + x}{x}\right) (\sqrt{1 - x^2} + 1 + x)} \end{align*}$

Hopefully, you can go from here, right? :x

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  • $\begingroup$ I can, the numerator tends to $-2$ and the denominator to $4$ using the fact about $arcsin{x}$ I mentioned in my post; thanks a lot. That limit after using the L'Hospital's rule for the first time just seemed so demotivating, yet it gives the right answer. $\endgroup$ – David May 24 '14 at 10:32
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    $\begingroup$ Just a thing to remember, when you are trying to get rid of a square root, it's conjugate will help you in most cases. After that you'll have a polynomial, and from here, it should be easy to eliminate all factors that tend to 0. That's basically what I do in the last step. $\endgroup$ – user49685 May 24 '14 at 10:38
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We write the Taylor series:

$$\frac{\ln(1+x)}{\arcsin x}=\frac{x-\frac{x^2}{2}+o(x^2)}{x+o(x^2)}=1-\frac{x}{2}+o(x)$$ hence we see now easily that the desired limit is $-\frac12$.

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