I would want to know how we can compute the following limit by using only fundamental limits.

$$\lim_{x \to a} \dfrac{a^x-x^a}{x-a},$$ where $a$ is a positive real number.

My idea was to use a substitution: $y=x-a$. We get $$\lim\limits_{y \to 0} \dfrac{a^aa^y-(y+a)^a}{y} =a^a\left[ \lim\limits_{y \to 0} \dfrac{a^y-1+1-(\frac{y+a}{a})^a}{y} \right] =a^a\left[ \ln a+\lim\limits_{y \to 0}\frac{1-(\frac{y+a}{a})^a}{y} \right]. $$

I am looking forward to read any tips on how I can continue from this point. Any help is appreciated.


3 Answers 3


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

where $f(x)=a^x$ and $g(x)=x^a$

  • $\begingroup$ No L'Hopital means no derivatives. $\endgroup$
    – user499203
    Nov 26, 2017 at 0:37
  • 4
    $\begingroup$ @ThePirateBay...No L'Hospital means no L'Hospital not no derivatives...;) ..Also it is not stated in the question that we must not use the definition of the derivative... $\endgroup$ Nov 26, 2017 at 1:48
  • 2
    $\begingroup$ I contend that this answer doesn’t invoke LH’s rule, because it makes no mention of the conditions required to apply the rule. Therefore, there’s no way that this could be a reformulation of the theorem. $\endgroup$ Nov 26, 2017 at 6:31

Your steps are good, now just apply well-known limit $$ \lim_{n \to 0} \frac{(n+1)^k - 1}{n} = k $$ Your last limit becomes $$\lim_{y\to0}\frac{(ya^{-1}+1)^a-1}{y}=aa^{-1}=1$$ Substituting it in your original expression $$a^a(\ln a+(-1))=a^a(\ln a-1)$$


hint: to conclude you shall use the limit $$ \lim_{x \to 0} \frac{(1+x)^a - 1}{x} = a $$ can you figure out how?

  • $\begingroup$ Thanks a lot. I completely forgot this result. $\endgroup$
    – Alchimist
    Nov 25, 2017 at 18:54

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