A) 0
B) 1
C) 2
D) \[\infty \]
Correct Answer: D
Solution :
\[\underset{x\to 1-}{\mathop{\lim }}\,\,\,\frac{1}{|\,\,1-x\,\,|}=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{1}{1-(1-h)}=\infty \] and \[\underset{x\to 1+}{\mathop{\lim }}\,\,\,\frac{1}{|\,\,1-x\,\,|}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{1+h-1}=\infty \] Hence \[\underset{x\to 1}{\mathop{\lim }}\,\,\frac{1}{|\,\,1-x\,\,|}=\infty .\]
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