A) 1
B) 0
C) e
D) \[(1/e)\]
E) \[\infty \]
Correct Answer: A
Solution :
\[\underset{x\to a}{\mathop{\lim }}\,\frac{{{a}^{x}}-{{x}^{a}}}{{{x}^{x}}-{{a}^{a}}}\] \[=\underset{x\to a}{\mathop{\lim }}\,\frac{{{a}^{x}}{{\log }_{e}}a-a{{x}^{a-1}}}{{{x}^{x}}(1+\log x)}\] (by L Hospitals rule) \[=\frac{{{a}^{a}}{{\log }_{e}}a-{{a}^{a}}}{{{a}^{a}}(\log a+1)}=\frac{{{\log }_{e}}a-1}{{{\log }_{e}}a+1}\] \[\therefore \] \[\underset{x\to a}{\mathop{\lim }}\,\frac{{{a}^{x}}-{{x}^{a}}}{{{x}^{x}}-{{a}^{a}}}=-1\] \[\therefore \] \[{{\log }_{e}}a-1=-{{\log }_{e}}a-1\] \[\Rightarrow \] \[2{{\log }_{e}}a=0\] \[\Rightarrow \] \[a=1\]
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