A) 0
B) 1
C) \[\frac{1}{e}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
[c] \[\underset{x\to e}{\mathop{\lim }}\,\frac{\log x-1}{x-e}\]which is in \[\frac{0}{0}\]form \[=\underset{x\to e}{\mathop{\lim }}\,\frac{\frac{1}{x}-0}{1-0}\] [By L. Hospital rule] Applying limit \[=\frac{1}{e}\]Hence the option is (c).You need to login to perform this action.
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