Find lim as x to 0+ (arcsin sqrt(sin x))/(sqrt(2x-x²))
Find $\displaystyle\lim_{x\to0^{+}}\frac{\arcsin\sqrt{\sin x}}{\sqrt{2x-x^{2}}}$.
Explanation of Method
Three nested functions, each equivalent to something simple.
Step 1: Check the indeterminate form
As $x\to0^{+}$: $\sin x\to0^{+}$, so $\sqrt{\sin x}\to0$ and $\arcsin\sqrt{\sin x}\to\arcsin0 = 0$. The denominator is $\sqrt{2\cdot0-0^{2}} = 0$. The form is $\dfrac00$. Note that the one-sided limit is forced on us: for $x\lt 0$ the radicand $\sin x$ is negative.
Step 2: Replace the numerator by an equivalent
Since $\arcsin t\sim t$ as $t\to0$, and here $t = \sqrt{\sin x}\to0$,
\[ \arcsin\sqrt{\sin x} \sim \sqrt{\sin x} \sim \sqrt{x}, \]
the last step because $\sin x\sim x$ and equivalence is preserved under the square root.
Step 3: Factor the denominator
\[ \sqrt{2x-x^{2}} = \sqrt{x\bigl(2-x\bigr)} = \sqrt{x}\,\sqrt{2-x}, \]
and $\sqrt{2-x}\to\sqrt2$, so $\sqrt{2x-x^{2}}\sim\sqrt2\,\sqrt{x}$.
Step 4: Form the ratio and evaluate
\[ \frac{\arcsin\sqrt{\sin x}}{\sqrt{2x-x^{2}}} \sim \frac{\sqrt{x}}{\sqrt2\,\sqrt{x}} = \frac{1}{\sqrt2}. \]
Step 5: Verification
\[\tfrac{1}{\sqrt2} = 0.7071068;\quad x = 0.01:\ 0.7100621;\quad x = 0.001:\ 0.7074015;\quad x = 10^{-4}:\ 0.7071362 \;\;\checkmark\]
\[\text{Exact expansion: } \frac{\arcsin\sqrt{\sin x}}{\sqrt{2x-x^{2}}} = \frac{\sqrt2}{2}\left(1+\frac{5x}{12}+\cdots\right) \;\;\checkmark\]
\[\text{Check at } x = 0.01:\ 0.7071068\bigl(1+0.0041667\bigr) = 0.7100531 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to0^{+}}\frac{\arcsin\sqrt{\sin x}}{\sqrt{2x-x^{2}}} = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}\,}\]
Remark
Remark. Each substitution is an equivalence at the origin, not an approximation carried through algebra of unequal orders, so the chain is legitimate: numerator and denominator are both of order $\sqrt{x}$ and the half power cancels. Differentiating instead would mean handling $\dfrac{\cos x}{2\sqrt{\sin x}\sqrt{1-\sin x}}$ against $\dfrac{1-x}{\sqrt{2x-x^{2}}}$ — correct, but several lines longer for the same $1/\sqrt2$. The expansion shows the approach is from above, at a rate of about $5x/12$, exactly as the numbers behave.