A) 1
B) ?1
C) 0
D) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
Applying L-Hospital?s rule, \[\underset{x\to 1}{\mathop{\text{lim}}}\,\,\,\frac{1+\log x-x}{1-2x+{{x}^{2}}}=\underset{x\to 1}{\mathop{\text{lim}}}\,\,\,\frac{\frac{1}{x}-1}{-2+2x}=\underset{x\to 1}{\mathop{\text{lim}}}\,\,\,\frac{1-x}{2x(x-1)}\] Again applying L-Hospital?s rule, \[\underset{x\to 1}{\mathop{\text{lim}}}\,\frac{-1}{4x-2}=-\frac{1}{2}\].
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