MCQ
If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$, then
- ✓$a = 1$
- B$a = 0$
- C$a = e$
- DNone of these
$ - 1 = \mathop {\lim }\limits_{x \to a} \,\,\frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = \mathop {\lim }\limits_{x \to a} \,\frac{{{a^x}{{\log }_e}a - a{x^{a - 1}}}}{{{x^x} + {a^a}{{\log }_e}a}}$
$ \Rightarrow \,\, - 1 = \frac{{{a^a}{{\log }_e}a - a\,.\,{a^{a - 1}}}}{{{a^a} + {a^a}{{\log }_e}a}} = \frac{{{{\log }_e}a - 1}}{{{{\log }_e}a + 1}}$.....$(i)$
Now $(i)$ is satisfied only when $a = 1.$
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