A) 1
B) \[-1\]
C) zero
D) \[-1/2\]
Correct Answer: D
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\,\frac{1+\log x-x}{1-2x+{{x}^{2}}}=\frac{0}{0}\] Applying L? Hospital rule \[\underset{x\to 1}{\mathop{\lim }}\,\frac{\frac{1}{x}-1}{-2+2x}=\underset{x\to 1}{\mathop{\lim }}\,\frac{1-x}{2x(x-1)}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{-(x-1)}{2x(x-1)}=-1/2\]
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