I managed to show with de L'Hospital that $$\lim_{x \to 0+} \sqrt[x]{x} = 0.$$
I calculated the first two derivatives and realised that it is getting more and more complicated due to the product rule and powers of logarithms to show that the limit $$\lim_{x \to 0+} \frac{d^k}{dx^k} \sqrt[x]{x} = 0$$
which some plots that I made are indicating.
I also thought about using $$\lim_{x \to 0+} \exp\left(-{\frac{1}{x}}\right)^{\ln(x)}$$ because inductively the limits of $\exp\left(- \frac{1}{x} \right)$ are a lot easier to calculate, though I didn't come up with a general formula and don't know how to use it properly with the given problem above.
A result that might be helpful, that I could proof, was that for all $\alpha >0$ it follows that $$ \lim_{x \to 0+} x^{\alpha} \, \ln(x) = 0.$$
Could you offer me any hints how to get to the desired result by induction? Is there an easy way to exchange the limit with the derivative?
Ideas so far: Let's restrict ourself to the compact intervall $[0,1]$. $f$ being a realvalued continous and bounded function defined on $(0,1]$ is uniform continous iff $f$ is can be continously extended on $[0,1]$. As this is valid for $k=0$, we exchange the first derivative with the limit and get that $\lim_{x \to 0+} (x^{\frac{1}{x}}) ' = 0$. As $(x^{\frac{1}{x}}) '$ is bounded, we can repeat the argument inductively.
The first two derivates are:
$$(x^{\frac{1}{x}}) ' = \exp \left( \tfrac{\ln(x)}{x} \right)\,\left( \frac{1-\ln(x)}{x^2} \right) = \sqrt[x]{x}\,\left( \frac{1-\ln(x)}{x^2} \right)$$ $$ (x^{\frac{1}{x}}) '' = \sqrt[x]{x} \, \left( \ln^2(x) - 3\,x + 2\, (x-1)\, \ln(x) + 1 \right) \, \frac{1}{x^4} $$
My answer is not complete, but I think it's a good start.
Let $h(x) =x^{1/x} $ and $f_k(x) =(h(x))^{(k)} =(e^{\ln(x)/x})^{(k)} $.
Note that $h'(x) =f_1(x) = h(x)\left( \frac{1-\ln(x)}{x^2} \right) $.
From the computations, it looks like $f_k(x) =h(x)g_k(x)x^{-2k} $ for some $g_k(x)$. In particular, $g_1(x) =1-\ln(x) $.
Then
$\begin{array}\\ f_{k+1}(x) &=(f_k(x))'\\ &=(h(x)g_k(x)x^{-2k})'\\ &=h'(x)g_k(x)x^{-2k}+h(x)g_k'(x)x^{-2k}+h(x)g_k(x)(x^{-2k})'\\ &=h(x)\left( \frac{1-\ln(x)}{x^2} \right)g_k(x)x^{-2k} +h(x)g_k'(x)x^{-2k} +h(x)g_k(x)(-2k)x^{-2k-1}\\ &=h(x)x^{-2k-2}\left( (1-\ln(x))g_k(x)+x^2g_k'(x)-2kxg_k(x)\right)\\ &=h(x)x^{-2k-2}\left( (1-\ln(x)-2kx)g_k(x)+x^2g_k'(x)\right)\\ \end{array} $
so if $g_{k+1}(x) =(1-\ln(x)-2kx)g_k(x)x+x^2g_k'(x) $, and we can show that $h(x)g_k(x)x^{-2k} \to 0$, we are done.
$h(x) =e^{\ln(x)/x} $ so, if $y = 1/x$, $h(y) =e^{-y\ln(y)} $ so $g_{k+1}(1/y) =(1+\ln(y)-2k/y)g_k(1/y)x+g_k'(1/y)/y^2 $.
This last might be better by looking when $y \to \infty$.
Anyway, that's all I have time for right now, so I'll stop,