Is there another way to compute $ \lim_{x \to 0^+} (\ln x)/{x^{1/9} } $? Im trying to compute the limit
$$ \lim_{x \to 0^+} \frac{ \ln x}{x^{1/9} } $$
Im trying l'hospitals it doesnt work. What trick do we need here?
 A: Without appealing to L'Hospital's Rule, we note that $\log(x)\le x-1$ for $x>0$.  
Then, we have
$$\begin{align}
\frac{\log(x)}{x^{1/9}}&\le \frac{x-1}{x^{1/9}}\\\\
&=x^{8/9}-x^{-1/9}\\\\
&\to -\infty \,\,\text{as}\,\,x\to 0^+
\end{align}$$
And we are done!

It is interesting to note that L'Hospital's Rule does indeed apply to the limit of the reciprocal, $\frac{x^{1/9}}{\log(x)}$. That is, 
$$\begin{align}
\lim_{x\to 0^+} \frac{x^{1/9}}{\log(x)}&=\lim_{x\to 0^+}\frac{\frac19 x^{−8/9}}{1/x}\\\\
&=\lim_{x\to 0^+} \frac19 x^{1/9}\\\\
&=0
\end{align}$$
Now, since for $0<x<1$, $\log(x)<0$, we immediately see that 
$$\lim_{x\to 0^+}\frac{\log(x)}{x^{1/9}}=-\infty$$
A: Use the substitution $x=y^9$, where $y>0$.
A: One may just write
$$
 \lim_{x \to 0^+} \frac{ \ln x}{x^{1/9} } = \lim_{x \to 0^+}\left( \frac1{x^{1/9} } \times   \ln x \right)=\frac1{0^+} \times (-\infty)=+\infty \times (-\infty)=-\infty.
$$
A: It is not an indeterminate form: it is: ‘ $\dfrac{-\infty}{0^+}=-\infty\times(+\infty)=-\infty$’.
