Let f ( x ) = 1 - x ( 1 + | 1 - x | ) | 1 - x | cos ( 1 1 - x ) , for x 1 . Then [2017]

Question

Let $f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} \cos (\frac{1}{1 - x}), for x \neq 1 .$ Then [2017]

Smart Achievers · Accepted Answer

(1, 4)

$Given: f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} \cos (\frac{1}{1 - x}), x \neq 1$

$\lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} \frac{1 - (1 - h) (1 + h)}{h} \cos (\frac{1}{h})$

$= \lim_{h \to 0} \frac{1 - 1 + h^{2}}{h} \cos (\frac{1}{h}) = \lim_{h \to 0} h \cos (\frac{1}{h}) = 0$

$\lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} \frac{1 - (1 + h) (1 + h)}{h} \cos (\frac{1}{h})$

$= \lim_{h \to 0} \frac{- 2 h - h^{2}}{h} \cos (\frac{1}{h}) = \lim_{h \to 0} (- 2 - h) \cos (\frac{1}{h})$

$= - 2 \times (some value oscillating between - 1 and 1)$

$∴$ $\lim_{x \to 1^{+}} f (x) does not exist.$

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