# What is the correct option of the following question?

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#### Explanation

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#### Explanation:

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Mar 26, 2018

${\lim}_{x \to {1}^{-}} \frac{1 - \sqrt{x}}{\arccos x} ^ 2 = \frac{1}{4}$

#### Explanation:

Substitute $x = \cos t$, $t = \arccos x$ so that:

${\lim}_{x \to {1}^{-}} t \left(x\right) = \arccos \left(1\right) = 0$

Then:

${\lim}_{x \to {1}^{-}} \frac{1 - \sqrt{x}}{\arccos x} ^ 2 = {\lim}_{t \to {0}^{+}} \frac{1 - \sqrt{\cos} t}{t} ^ 2$

Rationalize the numerator:

${\lim}_{x \to {1}^{-}} \frac{1 - \sqrt{x}}{\arccos x} ^ 2 = {\lim}_{t \to {0}^{+}} \frac{\left(1 - \sqrt{\cos} t\right) \left(1 + \sqrt{\cos} t\right)}{{t}^{2} \left(1 + \sqrt{\cos} t\right)}$

${\lim}_{x \to {1}^{-}} \frac{1 - \sqrt{x}}{\arccos x} ^ 2 = {\lim}_{t \to {0}^{+}} \frac{1 - \cos t}{{t}^{2} \left(1 + \sqrt{\cos} t\right)}$

Use now the well known trigonometric limit:

${\lim}_{\theta \to 0} \frac{1 - \cos \theta}{\theta} ^ 2 = \frac{1}{2}$

to have:

${\lim}_{x \to {1}^{-}} \frac{1 - \sqrt{x}}{\arccos x} ^ 2 = {\lim}_{t \to {0}^{+}} \frac{1 - \cos t}{t} ^ 2 \cdot {\lim}_{t \to {0}^{+}} \frac{1}{1 + \sqrt{\cos} t}$

${\lim}_{x \to {1}^{-}} \frac{1 - \sqrt{x}}{\arccos x} ^ 2 = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

graph{ (1-sqrtx)/(arccosx)^2 [-0.51, 1.99, -0.625, 0.625]}

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