First slide

Continuity

Question

$Let α \in R be such that the function f (x) = \{\begin{matrix} \frac{\cos^{- 1} (1 - {\{x\}}^{2}) \sin^{- 1} (1 - \{x\})}{{x} - {x}^{3}} & , x \neq 0 \\ α, & x = 0 \end{matrix}$ $is continuous at x = 0, where$ ${x} = x - [x], [x] i s t h e g r e a t e s t i n t e g e r l e s s t h a n o r$ $e q u a l t o x . T h e n :$

Moderate

A

$α = 0$
B

$no such α exists$
C

$α = \frac{π}{4}$
D

$α = \frac{π}{\sqrt{2}}$

Solution

$RHL = \lim_{x \to 0^{+}} \frac{\cos^{- 1} (1 - x^{2}) \sin^{- 1} (1 - x)}{x (1 - x^{2})} = \frac{π}{2} \lim_{x \to 0^{+}} \frac{\cos^{- 1} (1 - x^{2})}{x}$
$= \frac{π}{2} \lim_{x \to 0^{+}} \frac{- 1}{\sqrt{1 - {(1 - x^{2})}^{2}}} (- 2 x) (L'Hospital Rule)$
$= π \lim_{x \to 0^{+}} \frac{x}{\sqrt{2 x^{2} - x^{4}}} = π \lim_{x \to 0^{+}} \frac{1}{\sqrt{2 - x^{2}}} = \frac{π}{\sqrt{2}}$
$LHL = \lim_{x \to 0^{+}} \frac{\cos^{- 1} (1 - (1 + x)^{2}) \sin^{- 1} (- x)}{(1 + x) - (1 + x)^{3}} = \frac{π}{2} \lim_{x \to 0^{+}} \frac{\sin^{- 1} x}{(1 + x) [(1 + x)^{2} - 1]} = \frac{π}{2} \lim_{x \to 0^{+}} \frac{\sin^{- 1} x}{x^{2} + 2 x}$
$= \frac{π}{2} (\frac{1}{2}) = \frac{π}{4} As LHL \neq RHL so f (x) is not continuous at x = 0 .$
$Therefore no such α exists$

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