A) \[0\]
B) \[1\]
C) \[a\]
D) does not exist
Correct Answer: B
Solution :
\[\underset{x\to a}{\mathop{\lim }}\,\frac{\log (x-a)}{\log ({{e}^{x}}-{{e}^{a}})}\left( \frac{\infty }{\infty }form \right)\] Using L' Hospital's rule, we get \[=\underset{x\to a}{\mathop{\lim }}\,\frac{\frac{1}{x-a}}{\frac{{{e}^{x}}}{{{e}^{x}}-{{e}^{a}}}}=\underset{x\to a}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{a}}}{(x-a)({{e}^{x}})}\] Again, using L' Hospital's rule \[=\underset{x\to a}{\mathop{\lim }}\,\frac{{{e}^{x}}}{{{e}^{x}}(x-a)+{{e}^{x}}}=\frac{{{e}^{a}}}{{{e}^{a}}}=1\]
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