Question

For each \(t\in\mathbb{R}\), let \([t]\) be the greatest integer less than or equal to \(t\). Then \(\displaystyle\lim_{x\to1^+}\frac{(1-|x|+\sin|1-x|)\,\sin\left(\frac{\pi}{2}[1-x]\right)}{|1-x|\,[1-x]}\)

• A. equals \(-1\)
• B. equals 1
• C. equals 0
• D. does not exist

Hint:

For \(x\to1^+\), \(1-x\) is a small negative number, so \([1-x]=-1\); then expand the numerator.

Answer 1

The Correct Option is C

Let \(x=1+h\), \(h\to0^+\). Then \(|x|=1+h\), \(|1-x|=h\), \([1-x]=[-h]=-1\).
Numerator: \((1-(1+h)+\sin h)\sin\left(-\frac{\pi}{2}\right)=(\sin h-h)(-1)=h-\sin h\).
Denominator: \(h\cdot(-1)=-h\).
Limit \(=\lim_{h\to0^+}\frac{h-\sin h}{-h}=-\lim_{h\to0^+}\left(1-\frac{\sin h}{h}\right)=-(1-1)=0\).
Answer: (3) equals 0