Correct Answer - Option 3 : 1
Concept:
Indeterminate Forms: Any expression whose value cannot be defined, like \(\rm \frac00\), \(\rm \pm \frac{\infty}{\infty}\), 00, ∞0 etc.
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L'Hospital's Rule: For the differentiable functions f(x) and g(x), the \(\rm \lim_{x\to c} \frac{f(x)}{g(x)}\), if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the \(\rm \lim_{x\to c} \frac{f'(x)}{g'(x)}\) if it exists.
- For the indeterminate form ∞ - ∞, first rationalize by multiplying with the conjugate and then divide the terms by the highest power of the variable to get terms so that \(\rm \frac1{x}\) → 0 as x → ∞.
- For the indeterminate form \(\rm \frac00\), first try to rationalize by multiplying with the conjugate, or simplify by cancelling some terms in the numerator and denominator. Else, use the L'Hospital's rule.
Calculation:
\(\rm \lim_{x \to a} \frac{a^x -x^a}{x^x -a^a}\)
= \(\rm \frac{a^a - a^a}{a^a - a^a}\)
= \(\frac00\), an indeterminate form.
Applying L'Hospital's rule, we get:
\(\rm \lim_{x \to a} \frac{a^x -x^a}{x^x -a^a}\)
= \(\rm \lim_{x \to a} \frac{a^x\log a -ax^{a-1}}{x^x(\log x +1)}\)
= \(\rm \frac{a^a\log a -a\cdot a^{a-1}}{a^a(\log a +1)}\)
= \(\rm \frac{\log a -1}{\log a +1}\)
According to the question, \(\rm \frac{\log a -1}{\log a +1}\) = -1.
⇒ log a - 1 = - log a - 1
⇒ 2 log a = 0
⇒ a = 1.