| Then \[\underset{x\to 1+}{\mathop{\lim }}\,\,\frac{1-\left| x \right|+\sin \left| 1-x \right|)sin\left( \frac{\pi }{2}[1-x] \right)}{\left| 1-x \right|\,.\,[1-x]\,}\] |
A) equals -1
B) equals 1
C) equals 0
D) does not exist
Correct Answer: C
Solution :
\[\underset{x\to 1+}{\mathop{\lim }}\,\frac{(1-\left| x \right|+\sin \left| 1-x \right|)\sin \,\left( \frac{\pi }{2}[1-x] \right)}{\left| 1-x \right|.[1-x]}\] \[x=1+h\] \[=\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{(1-1-h\,\,\sin \,h)\,\sin \,\left( \frac{\pi }{2}(-1) \right)}{h(-1)}\] \[=\,\,\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{\sinh -h}{-h} \right)(-1)\] \[=\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h}{h}\,\,-1=1-1=0\]
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