Let $f(x)=\frac{1-x(1+|1-x|)}{|1-x|} \cos \left(\frac{1}{1-x}\right)$ for $x \neq 1$. Then

(A) $\lim_{x \rightarrow 1^{-}} f(x)=0$; (D) $\lim_{x \rightarrow 1^{+}} f(x)$ does not exist

(B) $\lim_{x \rightarrow 1^{-}} f(x)$ does not exist

(C) $\lim_{x \rightarrow 1^{+}} f(x)=0$

Limits — JEE Advanced · 2017 · 21 May 2017 · Paper 2 Question with Solution

Let

f (x) = \frac{1 - x ( 1 + ∣1 - x ∣ )}{∣1 - x ∣} cos (\frac{1}{1 - x})

for

x \neq = 1

. Then

Answer: (A), (D)

For

x < 1

: Let

x = 1 - h

where

h > 0

∣1 - x ∣ = h, 1 + ∣1 - x ∣ = 1 + h

f (x) = \frac{1 - ( 1 - h ) ( 1 + h )}{h} cos (\frac{1}{h}) = \frac{1 - ( 1 - h ^{2} )}{h} cos (\frac{1}{h}) = \frac{h ^{2}}{h} cos (\frac{1}{h}) = h cos (\frac{1}{h})

h \to 0^{+}

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

(since

∣ cos (1/ h) ∣ \leq 1

and

h \to 0

)

For

x > 1

: Let

x = 1 + h

where

h > 0

∣1 - x ∣ = h, 1 + ∣1 - x ∣ = 1 + h

f (x) = \frac{1 - ( 1 + h ) ( 1 + h )}{h} cos (\frac{1}{- h}) = \frac{1 - ( 1 + h ) ^{2}}{h} cos (- \frac{1}{h}) = \frac{- h ( 2 + h )}{h} cos (\frac{1}{h})

= - (2 + h) cos (\frac{1}{h})

h \to 0^{+}

cos (1/ h)

oscillates between -1 and 1, so the limit does not exist.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options