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EEJ MAIN Mathematics QUESTION #6953
Question 1

For each $t\in\mathbb{R}$, let $[t]$ denote the greatest integer $\leq t$. Evaluate: $\displaystyle\lim_{x\to1^+}\dfrac{(1-|x|+\sin|1-x|)\sin\!\left(\dfrac{\pi}{2}[1-x]\right)}{|1-x|\cdot[1-x]}$

  • Equals 1
  • Equals 0✔️
  • Equals $-1$
  • Does not exist
Correct Answer Explanation

As $x\to1^+$: let $h=x-1>0$, so $h\to0^+$. Then $|1-x|=h$, $[1-x]=[-h]=-1$.

$\sin\!\left(\dfrac{\pi}{2}[-h]\right)=\sin\!\left(\dfrac{\pi}{2}\cdot(-1)\right)=\sin(-\pi/2)=-1$

Numerator: $(1-|x|+\sin h)\cdot(-1)$. As $x\to1^+$: $|x|=x=1+h$, so $1-|x|=-h$.

Numerator $= (-h+\sin h)(-1) = h-\sin h$

Denominator: $h\cdot(-1)=-h$

Limit $= \dfrac{h-\sin h}{-h} = -(1-\dfrac{\sin h}{h})\to-(1-1)=\mathbf{0}$