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.
Solve this limit $\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}$
2
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calculus
limits
2 Answers
1
\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}
0
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 $y \to 0^{+}$