A) 0
B) 1
C) a
D) does not exist
E) \[-a\]
Correct Answer: B
Solution :
\[\underset{x\to a}{\mathop{\lim }}\,\frac{\log (x-a)}{\log ({{e}^{x}}-{{e}^{a}})}\] \[=\underset{x\to a}{\mathop{\lim }}\,\frac{\frac{1}{x-a}}{\frac{1}{{{e}^{x}}-{{e}^{a}}}{{e}^{x}}}\](by LHospitals rule) \[=\underset{x\to a}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{a}}}{{{e}^{x}}(x-a)}\] \[=\underset{x\to a}{\mathop{\lim }}\,\frac{{{e}^{x}}}{{{e}^{x}}(x-a)+{{e}^{x}}}\] (again by LHospitals rule) \[=\frac{{{e}^{a}}}{{{e}^{a}}}=1\]You need to login to perform this action.
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