# limit of a function involving $\arcsin{x}$ and $\ln{(1+x)}$

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

• just rewrite it with one numerator and one denominator and apply L'Hospital's rule twice. – Apurv May 24 '14 at 10:13
• @Apurv I thought I might get away from taking the derivative of that, but it actually gives the result, thanks. – David May 24 '14 at 10:33

\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*}
• 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. – David May 24 '14 at 10:32
$$\frac{\ln(1+x)}{\arcsin x}=\frac{x-\frac{x^2}{2}+o(x^2)}{x+o(x^2)}=1-\frac{x}{2}+o(x)$$ hence we see now easily that the desired limit is $-\frac12$.