$\lim_{x\to a}\frac{a^x-x^a}{x-a}$ without the L'Hopital's rule

Question

This limit is proposed to be solved without using the L'Hopital's rule or Taylor series: $$ \lim_{x\to a}\frac{a^x-x^a}{x-a}, $$ where $a>0$. I know how to calculate this limit using the L'Hopital's rule: $$ \lim_{x\to a}\frac{a^x-x^a}{x-a}= \lim_{x\to a}\frac{a^x\ln a-ax^{a-1}}{1}=a^a\ln a- a^a, $$ but I have no idea how to calculate this limit without using derivatives.

Answer 1

The limit is same thing to write as, $$\lim_{k\to 0}\frac{a^{a+k}-(a+k)^{a}}{k}$$ Now, I will use some non standard Analysis $$st\left(\frac{a^{a+\epsilon}-(a+\epsilon)^{a}}{\epsilon}\right)$$ Where $\epsilon$ denotes an infinitesimal. And $st(...)$ for standard part function. In Euler's famous book, "Indroductio in analysin infnitorum" chapter 7 he derives the formula that $a^{\epsilon}=1+\ln(a)\epsilon$ Putting this in required limit and using the binomial theorem, $$st\left(\frac{1}{\epsilon}\left(a^{a}+a^{a}\epsilon\ln a-a^{a}-\epsilon a^{a}-\binom{a}{2}a^{a-2}(\epsilon)^{2}-...\right)\right)$$ Notice that $a^{a}$ cancels off each other. The infinitesimal in the denominator cancels with epsilon with the numerator with $a^{a}(\ln a-1)$. All other terms will get $\epsilon^{n}, n\in\mathbb{N}$ as their coefficients. So the required answer is, $a^{a}(\ln a-1)$. $$\lim_{x\to a}\frac{a^x-x^a}{x-a}=a^{a}(\ln a-1)$$

Answer 2

$$ L=\lim_{x\to a}\frac{a^x-x^a}{x-a}=\lim_{h\to 0}\frac{a^{a+h}-(a+h)^a}{h}=\lim_{h\to 0}a^a\frac{a^h-(1+h/a)^a}{h}.$$ Use binomial expansion $(1+z)^k=1+kz+k(k-1)z^2/2!+...$ and $\lim_{z\to 0} \frac{a^z-1}{z}=\ln a$ $$\implies L=a^a\lim_{h \to 0}\frac{a^h-(1+h+ h^2a(a-1)/a^2}{h}=a^a[\lim_{h\to 0}\frac{a^h-1}{h}-1]=a^a (\log a-1)$$

Answer 3

You do not need L'Hopital or Taylor series here. If you rewrite the numerator as $(a^x-a^a)-(x^a-a^a)$ and use the definition of the derivative, you have

$$\lim_{x\to a}{a^x-x^a\over x-a}=\lim_{x\to a}{a^x-a^a\over x-a}-\lim_{x\to a}{x^a-a^a\over x-a}=f'(a)-g'(a)$$

where $f(x)=a^x$ and $g(x)=x^a$ have derivatives $f'(x)=a^x\ln a$ and $g'(x)=ax^{a-1}$, respectively, hence

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

This may look like L'Hopital, but it's not. Indeed, anytime you have a L-Hopital-like problem, $\lim_{x\to a}f(x)/g(x)$ with $g'(a)\not=0$, you don't need to invoke L'Hopital to evaluate the limit; L'Hopital can be a convenience in such settings, but it is really only needed when $\lim_{x\to a}f'(x)$ and $\lim_{x\to a}g'(x)$ are also both $0$.