I have encountered this problem and I don't know how to approach it, mainly because we can't use L'Hospital's Rule. The limit is as follows:

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

Thanks for any suggestions.

edit: Solved, thanks for the responses.

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    $\begingroup$ This is just the derivative of $f(x)=x^x$ at $a$, to evaluate it write $f(x)=e^{x\log x}.$ $\endgroup$ – omar Mar 4 '14 at 20:05
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    $\begingroup$ What are you allowed to use? Have you tried anything? $\endgroup$ – robjohn Mar 4 '14 at 20:07
  • $\begingroup$ @robjohn Well you can use esentially anything besides what I mentioned, I mean using some suitable algebra to identify some known limits including e in this case, such as: $\lim_{x\to 0} {e^x - 1 \over x}=1$. $\endgroup$ – David Mar 4 '14 at 20:20
  • $\begingroup$ as @omar explained,use $f'(a)=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$,by setting $x-a=h\implies \lim_{h\to 0}\dfrac{(a+h)^{(a+h)}-a^a}{h} =f'(a) $ with $f(x)=x^x$ $\endgroup$ – Jonas Kgomo Mar 4 '14 at 20:23
  • $\begingroup$ People will see that you've accepted an answer. There is no need to change the title to include [Solved]. Besides, someone may come up with a better solution; adding [Solved] to the title might discourage them. $\endgroup$ – robjohn Mar 4 '14 at 22:00

Hint 1: $$ \begin{align} \frac{e^{x\log(x)}-e^{a\log(a)}}{x-a} &=e^{a\log(a)}\frac{e^{x\log(x)-a\log(a)}-1}{x-a}\\ &=e^{a\log(a)}\frac{e^{\color{#C00000}{x\log(x)-a\log(a)}}-1}{\color{#C00000}{x\log(x)-a\log(a)}}\frac{x\log(x)-a\log(a)}{x-a} \end{align} $$ Hint 2: $$ \begin{align} \frac{x\log(x)-a\log(a)}{x-a} &=\frac{x(\log(x)-\log(a))}{x-a}+\log(a)\\ &=\frac xa\frac{\color{#C00000}{\log(x)-\log(a)}}{e^{\color{#C00000}{\log(x)-\log(a)}}-1}+\log(a) \end{align} $$ Apply $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$.

Although you commented that you could use $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$, here is a proof of this fact using the inequality $1+x\le e^x$. This inequality is shown using Bernoulli's Inequality in this answer.

For $|x|\lt1$, $1+x\le e^x$ and $1-x\le e^{-x}$. Therefore, $$ 1+x\le e^x\le\frac1{1-x} $$ Subtracting $1$ and dividing by $x$ yields $$ \begin{array}{} 1\le\frac{e^x-1}{x}\le\frac1{1-x}&\text{when }x\gt0\\ 1\ge\frac{e^x-1}{x}\ge\frac1{1-x}&\text{when }x\lt0 \end{array} $$ That is, when $|x|\lt1$, $$ \frac{e^x-1}{x}\text{ is between }1\text{ and }\frac1{1-x} $$ By the Squeeze theorem, $$ \lim_{x\to0}\frac{e^x-1}{x}=1 $$

  • $\begingroup$ Nice way to go (+1) $\endgroup$ – user 1357113 Mar 5 '14 at 15:38

No we can use Lopital's rule so

If $y=x^x$ and $x\gt0$ then $\ln(y)=\ln(x^x)$ Use properties of logarithmic functions to expand the right side of the above equation as follows. $$ \ln(y)=x\ln(x) $$ We now differentiate both sides with respect to x, using chain rule on the left side and the product rule on the right. $$ y'(1/y)=\ln(x) + x(1/x)=\ln(x) + 1 \text{, where }y'=\mathrm{d}y/\mathrm{d}x $$ Multiply both sides by $y$ $$ y'=(\ln(x) + 1)\,y $$ Substitute $y$ by $x^x$ to obtain $$ y'=(\ln(x) + 1)\,x^x $$ then $$ \lim_{x\to a} {x^x-a^a \over x-a}=\lim_{x\to a} {(\ln(x) + 1)\,x^x \over 1}=(\ln(a) + 1)\,a^a $$


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