$\displaystyle \lim_{x \to 0+} \frac{d^k}{dx^k} \sqrt[x]{x}$ for all $k \in \mathbb{N}$ and $x>0$ 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:
$$(x^{\frac{1}{x}}) ' = \exp \left( \tfrac{\ln(x)}{x} \right)\,\left( \frac{1-\ln(x)}{x^2} \right) = \sqrt[x]{x}\,\left( \frac{1-\ln(x)}{x^2} \right)$$
$$ (x^{\frac{1}{x}}) ''  =  \sqrt[x]{x} \, \left( \ln^2(x) - 3\,x + 2\, (x-1)\, \ln(x) + 1  \right) \, \frac{1}{x^4} $$
 A: 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,
A: A closed form can always be found by exploiting composition and using Faá di Brunos Formula
$$\frac{d^n}{dx^n} f(g(x))
 = \sum_{\substack{(m_1, \ldots, m_n): \\ \sum_{i = 1}^{n} i \cdot m_i = n}} \frac{n! \cdot f^{\left(\sum_{i = 1}^{n} m_i\right)} \left( g(x) \right)}{\prod_{i = 1}^{n} m_i! \cdot i !^{m_i}}  \prod_{j = 1}^{n} \left( g^{(j)}(x) \right)^{m_j}.
$$
