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Please help me solving $\displaystyle\lim_{x\to a}\frac{a^{a^{x}}-{a^{x^{a}}}}{a^x-x^a}$

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This problem is creepy! –  rschwieb Sep 12 '12 at 17:30
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And now for $$ \lim_{x \to a} \frac{a^{x^{a^x}} - x^{a^{x^a}}}{a^x - x^a} \, . $$ –  Hans Engler Sep 12 '12 at 18:28
    
@HansEngler where did you find this? –  yiyi Sep 13 '12 at 0:37

5 Answers 5

up vote 3 down vote accepted

$\displaystyle\lim_{x\to a}\frac{a^{a^{x}}-{a^{x^{a}}}}{a^x-x^a}$=$\displaystyle\lim_{x\to a}\frac{a^{x^{a}}({a^{a^{x}-x^a}}-1)}{a^x-x^a}$=$a^{a^{a}}\displaystyle\lim_{x\to a}\frac{{a^{a^{x}-x^a}-1}}{a^x-x^a}$=$|\displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=\ln a|$=$a^{x^{a}}\ln a$.

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For the denominator, you can write $x^a-a^x=x^a-a^a+a^a-a^x$ and use the derivative of each.

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The ratio is $R(x)=\dfrac{u(t)-u(s)}{t-s}$ with $u(z)=a^z$, $t=a^x$ and $s=x^a$. When $x\to a$, $t\to a^a$ and $s\to a^a$ hence $R(x)\to u'(a^a)$. Since $u(z)=\exp(z\log a)$, $u'(z)=u(z)\log a$. In particular, $u'(a^a)=u(a^a)\log a$. Since $u(a^a)=a^{a^a}$, $\lim\limits_{x\to a}R(x)=a^{a^a}\log a$.

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Let $f(u)=a^u$. Then the expression inside the limit is $\frac{f(z)-f(y)}{z-y}$ where $z=a^x$ and $y=x^a$.

So, by the mean value theorem, this expression is $f'(c)$ for some $c$ between $z$ and $y$.

But as $x\to a$, both $z$ and $y$ approach $a^a$, and therefore $c$ approaches $a^a$. Since $f'(u)$ is continuous, this means that the limit must be $f'(a^a)$.

So you just need to know $f'(u) = a^u\log a$.

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Since this is calculus why not try with L'Hospital?

$$\lim_{x\to a}\frac{a^{a^x}-a^{x^a}}{a^x-x^a}=\lim_{x\to a}\frac{a^xa^{a^x}\log^2a-ax^{a-1}\log a\cdot a^{x^a}}{a^x\log a-ax^{a-1}}=$$

$$=\frac{a^aa^{a^a}\log^2a-a^a\log a\cdot a^{a^a}}{a^a\log a-a^a}=\frac{a^{a^a}\log^2a-\log a\cdot a^{a^a}}{\log a -1}=a^{a^a}\log a$$

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