2
$\begingroup$

I am trying to figure out how to make this limit, even with the hopital. I've tried using hopital two times, but the situation 0/0 is still there. I've tried to solve it using wolfram, but I don't the solution. Even with rationalization + hopital nothing comes out. $$\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}$$ I wonder if there is some way to solve it, and would really appreciate any suggestion.

$\endgroup$
1
$\begingroup$

\begin{align} \lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}&=\lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1-4x^2}}\frac{2\sqrt{x-2x^2}}{1-4x}, \text{ L'hopital}\\ &= \lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1-2x}\sqrt{1+2x}}\frac{2\sqrt{x}\sqrt{1-2x}}{1-4x}\\ &=\lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1+2x}}\frac{2\sqrt{x}}{1-4x}\\ &=\frac{2}{\sqrt{2}}\frac{2\sqrt{\frac{1}{2}}}{1-2} \\&=-2\end{align}

$\endgroup$
0
$\begingroup$

Put $y = \sqrt{x-2x^2} \implies -2x^2+x -y^2 = 0\implies x = \dfrac{1+ \sqrt{1-8y^2}}{4}$. Can you substitute this into the arcsin and proceed to L'hospitale from here...with notice that this means $y \to 0^{+}$

$\endgroup$

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.