How to determine a limit without L'Hospital's rule

Question

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How to solve this type of limit without L'Hospital rule.

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

Answer 1

Hint: \begin{equation}a^x - a^a + a^a - x^a\end{equation}

Answer 2

Rewrite this quotient as $$\frac{a^x-x^a}{x-a}=\frac{a^x-a^a}{x-a}-\frac{x^a-a^a}{x-a},$$ which is the difference of two rates of variation. Can you proceed?

Answer 3

Taylor-expand $a^x$ and $x^a$ around $x=a$, $$a^x = a^a + a^a\ln a \>(x-a), \>\>\>\>\> x^a = a^a + a^a ( x-a)$$

Then,

$$\lim_{x\to a}\frac{a^x-x^a}{x-a}=\lim_{x\to a}\frac{a^a\ln a (x-a)-a^a ( x-a)}{x-a}=a^a(\ln a-1)$$

Answer 4

To make life easier, let $x=a+y$ $$\lim_{x\to a}\frac{a^x-x^a}{x-a}=\lim_{y\to 0}\frac{a^{a+y}-(a+y)^a}{y}$$ Factor out $a^a$ to make $$a^{a+y}-(a+y)^a=a^a\left(a^y-\left(1+\frac{y}{a}\right)^a \right)$$ Now $$a^y=e^{y \log(a)}=1+y \log (a)+\frac{1}{2} y^2 \log ^2(a)+O\left(y^3\right)$$ $$b=\left(1+\frac{y}{a}\right)^a\implies \log(b)=a \log\left(1+\frac{y}{a}\right)=y-\frac{y^2}{2 a}+O\left(y^3\right)$$ $$b=e^{\log(b)}=1+y+\frac{1}{2} \left(1-\frac{1}{a}\right) y^2+O\left(y^3\right)$$ $$a^y-\left(1+\frac{y}{a}\right)^a=y (\log (a)-1)+\frac{1}{2} y^2 \left(\frac{1}{a}+\log ^2(a)-1\right)+O\left(y^3\right)$$

$$\frac{a^{a+y}-(a+y)^a}{y}=a^a \left(\log (a)-1+\frac{1}{2} y \left(\frac{1}{a}+\log ^2(a)-1\right)+O\left(y^2\right) \right)$$ which not only shows the limit but also how it is approached.