A) \[1\]
B) \[-1\]
C) \[0\]
D) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{1+\log x-x}{1-2x+{{x}^{2}}}\] Since, the limit is of the form \[\left( \frac{0}{0} \right)\] so using L' Hospital rule, we get \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{\frac{1}{x}-1}{-2+2x}\] \[\left( \because \,\,form\frac{0}{0} \right)\] Again, using L' Hospital rule, \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{-\frac{1}{{{x}^{2}}}}{2}\] \[=\frac{-1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
