For each t ∈R, let [t ] be the greatest integer less than or equal to t. Then, limx 1_+ 0.05em (1- |x |+ sin ⁡ |1-x | ) sin ⁡( 2 [1-x ]) |1-x | [1-x ]

Question

Accepted Answer

Correct answer: B. (b) lim_x 1^+ 0.05em (1- |x |+ sin ⁡ ( |1-x | ) sin ⁡( 2 [1-x ])) |1-x | [1-x ] = limh 0 0.05em (1- |1+h |+ sin ⁡ ( |1-1-h | ) sin ⁡( 2 [1-1-h ])) |1-1-h | [1-1-h ] = limh 0 0.05em (1-1-h+ sinh ) sin ⁡( 2 (-1 ))h ( [0-h ] ) = limh 0 0.05em (-h+ sin ⁡h ) sin ⁡(- 2)h (-1 )=0

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

Solution

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