For each $t \in R$, let $[t]$ be the greatest integer less than or equal to $t$. Then, $\lim_{x \rightarrow 1+} \frac{(1-|x|+\sin |1-x|) \sin \left(\frac{\pi}{2}[1-x]\right)}{|1-x||1-x|}$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Limits

For each

t \in R

, let

[t]

be the greatest integer less than or equal to

t

. Then,

lim_{x \to 1 +} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ] )}{∣1 - x ∣∣1 - x ∣}

Answer: (B)

Approaching from

x \to 1^{+}

, write

x > 1

∣1 - x ∣ = x - 1

and

[1 - x] = - 1

, since

1 - x

is a small negative number. The sine factor becomes

sin (- \frac{π}{2}) = - 1

, and the expression reduces to

\frac{- ( 1 - x + sin ( x - 1 ))}{x - 1} = \frac{- ( x - 1 ) + sin ( x - 1 )}{x - 1} = - 1 + \frac{sin ( x - 1 )}{x - 1}

. As

x \to 1^{+}

the ratio

\frac{sin ( x - 1 )}{x - 1} \to 1

, so the limit is

- 1 + 1 = 0

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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