A) \[1\]
B) \[0\]
C) \[e\]
D) \[(1/e)\]
Correct Answer: A
Solution :
Given that,\[\underset{x\to a}{\mathop{\lim }}\,\frac{{{a}^{x}}-{{x}^{a}}}{{{x}^{x}}-{{a}^{a}}}=1\] Using L? Hospital?s rule \[\Rightarrow \] \[\underset{x\to a}{\mathop{\lim }}\,\frac{{{a}^{x}}{{\log }_{e}}a-{{x}^{a-1}}}{(1+{{\log }_{e}}x)-0}=-1\] \[\Rightarrow \] \[\frac{{{a}^{a}}{{\log }_{e}}a-{{a}^{a}}}{{{a}^{a}}({{\log }_{e}}a+1)}=\frac{{{\log }_{e}}a-1}{{{\log }_{e}}a+1}=-1\] \[\Rightarrow \] \[{{\log }_{e}}a-1=-{{\log }_{e}}a-1\] \[\Rightarrow \] \[2{{\log }_{e}}a=0\] \[\Rightarrow \] \[a=1\]
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