Question

If \(\lim\limits_{x \rightarrow a}\frac{(a^x-x^a)}{x^x-a^a}\) = -1 then, what is the value of a?

(a) 1

(b) 2

(c) 3

(d) 4

The question was posed to me by my college director while I was bunking the class.

Origin of the question is First Order Derivative in section Limits and Derivatives of Mathematics – Class 11

Answer 1

Right choice is (a) 1

For explanation: Let, y = x^x

Thus, log y = x log x

Differentiating both sides with respect to x, we get,

1/y dy/dx = (x*1/x) + log x

=>dy/dx = y(1 + log x)

Or, dx^x/dx = x^x(1 + log x)

Using, L’Hospital’s rule,

\(\lim\limits_{x \rightarrow a}\frac{(a^x-x^a)}{x^x-a^a}\) = \(\lim\limits_{x \rightarrow a}\frac{(a^x*log⁡a-x^{a-1})}{x^x(1+logx)}\)

= \(\lim\limits_{x \rightarrow a}\frac{(a^a*log⁡a-a^{a})}{a^a(1+loga)}\)

= (log a – 1)/(log a + 1)

As per the question,

(log a – 1)/(log a + 1) = -1

Or, (log a – 1) = -log a – 1

Or, 2 log a = 0

Or, log a = 0

So, a = 1