# lim x->oo e^x((2^(x^n))^(1/e^x)-(3^(x^n))^(1/e^x))/x^n , n belongs to N, is equal to ?

Jun 24, 2017

${\log}_{e} \left(\frac{2}{3}\right)$

#### Explanation:

Calling $u \left(x\right) = {e}^{x}$ and $v \left(x\right) = {x}^{n}$ we have

${e}^{x} \frac{{\left({2}^{{x}^{n}}\right)}^{\frac{1}{e} ^ x} - {\left({3}^{{x}^{n}}\right)}^{\frac{1}{e} ^ x}}{x} ^ n \equiv \frac{u \left(x\right)}{v \left(x\right)} \left({2}^{\frac{v \left(x\right)}{u \left(x\right)}} - {3}^{\frac{v \left(x\right)}{u \left(x\right)}}\right)$ and now making $\frac{u \left(x\right)}{v \left(x\right)} = \xi \left(x\right)$ we have

${\lim}_{x \to \infty} {e}^{x} \frac{{\left({2}^{{x}^{n}}\right)}^{\frac{1}{e} ^ x} - {\left({3}^{{x}^{n}}\right)}^{\frac{1}{e} ^ x}}{x} ^ n \equiv {\lim}_{x \to \infty} \xi \left(x\right) \left({2}^{\frac{1}{\xi \left(x\right)}} - {3}^{\frac{1}{\xi \left(x\right)}}\right)$

but $\xi \left(x\right) \to \infty$ as $x \to \infty$ so

${\lim}_{x \to \infty} {e}^{x} \frac{{\left({2}^{{x}^{n}}\right)}^{\frac{1}{e} ^ x} - {\left({3}^{{x}^{n}}\right)}^{\frac{1}{e} ^ x}}{x} ^ n \equiv {\lim}_{y \to \infty} \left(\frac{{2}^{\frac{1}{y}} - {3}^{\frac{1}{y}}}{\frac{1}{y}}\right) = {\log}_{e} \left(\frac{2}{3}\right)$

NOTE:

${\lim}_{y \to \infty} \left(\frac{{2}^{\frac{1}{y}} - {3}^{\frac{1}{y}}}{\frac{1}{y}}\right) \equiv {\lim}_{z \to 0} \frac{{2}^{z} - {3}^{z}}{z} = {\log}_{e} \left(\frac{2}{3}\right)$

because defining $f \left(z\right) = {2}^{z} - {3}^{z}$ we have

${\lim}_{h \to 0} \frac{f \left(0 + h\right) - f \left(0\right)}{h} = {\lim}_{h \to 0} \frac{\left({2}^{h} - {3}^{h}\right) - \left(1 - 1\right)}{h} = f ' \left(0\right)$ and

$f ' \left(z\right) = {2}^{z} {\log}_{e} 2 - {3}^{z} {\log}_{e} 3$ and then

$f ' \left(0\right) = {\log}_{e} \left(\frac{2}{3}\right)$