Finding limits $l=\lim_{x \rightarrow a}\frac{x^x-a^x}{x-a} $ & $m= \lim_{x \rightarrow a}\frac{a^x-x^a}{x-a} $.

Solution 1:

Hint for the first: Writing $h=x-a\to 0$, you can rewrite $$\begin{align} \frac{x^x-a^x}{x-a} &= \frac{e^{x\ln x}-e^{x\ln a}}{x-a} = \frac{e^{(a+h)\ln (a+h)}-e^{(a+h)\ln a}}{h} \end{align}$$ After doing this change towards $h\to 0$, if you are allowed to use Taylor series I would advocate (for both limits) to use Taylor expansions. The limits will then come quite easily. For instance, here is a derivation:

$$\begin{align}\frac{e^{(a+h)\ln (a+h)}-e^{(a+h)\ln a}}{h} &=\frac{e^{a(\ln a + \ln(1+\frac{h}{a})) + h(\ln a + \ln(1+\frac{h}{a})}-e^{a\ln a}e^{h\ln a}}{h} \\&=\frac{e^{a\ln a + a\cdot\frac{h}{a} + o(h) + h\ln a}-e^{a\ln a}e^{h\ln a}}{h} \\&= e^{a\ln a}\frac{e^{ h + \ln a + o(h)}-e^{h\ln a}}{h} \\&= e^{a\ln a}\frac{1+ h(1+\ln a) + o(h)- 1 - h\ln a}{h} \\&= e^{a\ln a}\frac{h + o(h)}{h} = e^{a\ln a}(1+o(1))\end{align}$$

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For the second: $$\begin{align} \frac{a^x-x^a}{x-a} &= \frac{e^{x\ln a}-e^{a\ln x}}{x-a} = \frac{e^{a\ln a+h\ln a}-e^{a\ln a + a\ln(1+\frac{h}{a})}}{h} \\ &=e^{a\ln a}\frac{e^{h\ln a}-e^{a(\frac{h}{a} + o(h))}}{h} = e^{a\ln a}\frac{e^{h\ln a}-e^{h + o(h)}}{h}\\ &= e^{a\ln a}\frac{1+h\ln a +o(h)-(1+h +o(h))}{h}\\ &= e^{a\ln a}\frac{h(\ln a - 1) +o(h)}{h} \\ &= e^{a\ln a}\left(\ln a - 1 + o(1)\right) \end{align}$$

Last part: If $m=\ell$, then $$e^{a\ln a}(\ln a -1) = e^{a\ln a}$$ Simplifying by $e^{a\ln a}$ and resolving the remaining equation will give you $a$.

Solution 2:

Hint for the first limit: I suppose that $a>0$. Put $g(x)=x^x=\exp(x\log x)$, and $h(x)=a^x$. Your expression is $\displaystyle \frac{g(x)-g(a)}{x-a}-\frac{h(x)-h(a)}{x-a}$