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Evaluate: $\displaystyle\lim_{x\to1^-}\dfrac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}}$
Let $\sin^{-1}x = \pi/2-\epsilon$ where $\epsilon\to0^+$ as $x\to1^-$. Then $x=\sin(\pi/2-\epsilon)=\cos\epsilon\approx1-\epsilon^2/2$, so $1-x\approx\epsilon^2/2$.
$\sqrt{\pi}-\sqrt{2(\pi/2-\epsilon)}=\sqrt{\pi}-\sqrt{\pi-2\epsilon}=\sqrt{\pi}\left(1-\sqrt{1-\frac{2\epsilon}{\pi}}\right)\approx\sqrt{\pi}\cdot\dfrac{\epsilon}{\pi}=\dfrac{\epsilon}{\sqrt{\pi}}$
$\sqrt{1-x}\approx\sqrt{\epsilon^2/2}=\epsilon/\sqrt{2}$
Limit $=\dfrac{\epsilon/\sqrt{\pi}}{\epsilon/\sqrt{2}}=\sqrt{\dfrac{2}{\pi}}$
A body moves eastward at $30\text{ m/s}$. After 10 s its velocity becomes $40\text{ m/s}$ northward. The magnitude of average acceleration is:
Change in velocity: $\Delta\vec{v} = 40\hat{j} - 30\hat{i}$
$|\Delta v| = \sqrt{30^2 + 40^2} = \sqrt{900+1600} = \sqrt{2500} = 50\text{ m/s}$
Average acceleration $= \dfrac{50}{10} = \mathbf{5\text{ m/s}^2}$
Given that $0 < x < y < 1$, compare the two quantities:
Quantity A: $1 - y$
Quantity B: $y - x$
We know $0 < x < y < 1$. Let's test specific values to see if the relationship is fixed.
Example 1: $x = 0.1,\ y = 0.2$.
Quantity A: $1 - 0.2 = 0.8$. Quantity B: $0.2 - 0.1 = 0.1$. A > B.
Example 2: $x = 0.1,\ y = 0.9$.
Quantity A: $1 - 0.9 = 0.1$. Quantity B: $0.9 - 0.1 = 0.8$. B > A.
Since the relationship changes depending on the values of $x$ and $y$, it cannot be determined from the information given.
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