Limit $ \lim_{x \to a}\left(\frac{a^x - x^a}{x-a}\right)$

Question

Need to obtain

$$ \lim_{x \to a}\left(\frac{a^x - x^a}{x-a}\right) $$

I used $ t = x - a$ to obtain $$\lim_{t \to 0}\left(\frac{a^a(a^t - 1)}{t}\right) = a^a\ln(a)$$

as I though asymptotically $ (a+t)^a = a^a $ as t approaches 0.

However, the answer is $ a^a\ln\left(\frac{a}{e}\right) $

Note: I will do not accept any answers which use Hospital rule (I'm in the part of the course there this is not covered, so this can be solved other way). Please answer in terms which math beginner knows.

Answer 1

Substitute

$$t:=x-a\,,\;\;\text{so that}\;x\to a\implies t\to0$$

and from here

$$\frac{a^{t+a}-(t+a)^a}t=\frac{a^a\left(a^t-\left(1+\frac ta\right)^a\right)}t=$$

$$=a^a\left(\frac{a^t-1}t-\frac{\left(1+\frac ta\right)^a-1}t\right)\xrightarrow[t\to0]{}a^a\left(\log a-1\right)$$

since (hope the following is also a basic, elementary limit)

$$\frac{\left(1+\frac ta\right)^a-1}t\xrightarrow[t\to0]{}1$$

Answer 2

Not sure what exactly the asker assumes. I'll assume that the derivatives are known.

First observe that the expression you have is:

$$ \frac{a^x -a^a}{x-a} -\frac{x^a -a^a}{x-a} $$

Now use definition of the derivative to conclude.

Specifically, the first summand tends to the derivative of $a^x $ at $x=a$, equal to $a^a \ln a$. The summand tends to the derivative of $x^a$ at $x=a$, equal to $a a^{a-1}=-a^a$. Therefore the limit is the difference between these two derivatives:

$$ a^a (\ln a -1) = a^a \ln (\frac a e ).$$