Evaluation of undefined limit I am supposed to evaluate this limit. 
$$\lim_{x\rightarrow 0} \, \frac{\sqrt[3]{x} \ln(\ln x)}{\sqrt[3]{(2x+3)\ln x}}$$
I tried to solve it as two limits, in the way that:
$$\lim_{x\rightarrow 0} \, \frac{\sqrt[3]{x}}{\sqrt[3]{2x+3}}$$
$$\lim_{x\rightarrow 0} \, \frac{\ln(\ln x)}{\sqrt[3]{\ln x}}$$
so that the first one is zero, but the second one is not difined for zero.
Can anyone help me to continue? 
Thanks.
 A: Maybe this is how you can see that the limit is equal to $0$: 
$$ \lim_{x \to 0} \frac{ \sqrt[3]{x} \log( \log(x)) }{ \sqrt[3]{2x + 3} \log(x)   } = 
  \frac{\lim_{x \to 0} \sqrt[3]{x} \log( \log(x)) }{ \lim_{x \to 0} \sqrt[3]{2x + 3} \log(x)   } = \frac{\lim_{x \to 0} \sqrt[3]{x} \log( \log(x)) }{ \lim_{x \to 0} \sqrt[3]{2x + 3} \lim_{x \to 0}  \log(x)   } = \frac{\lim_{x \to 0} \sqrt[3]{x} \log( \log(x)) }{  \sqrt[3]{3} \cdot (-\infty) } = 
=  \frac{\lim_{x \to 0} \sqrt[3]{x} \log( \log(x)) }{   (-\infty) } 
= \frac{ \sqrt[3]{\lim_{x \to 0} x \log^3( \log(x))} }{   (-\infty) } 
 = ^{L'Hospital} \frac{0}{-\infty} = 0 
 $$
A: This limit is bad -- $\ln\ln(x)$ doesn't exist when $x$ is close to $0$. Thus the function itself is undefined in the neighbourhood of $0$ (specifically, undefined when $x<1$, since $\ln(x) \le 0$ here so $\ln \ln x$ is undefined) so you can't say anything about this limit. Here's a picture:

A: Here I assume the question is to find
$$\tag 1 \lim_{x\rightarrow 0^+} \, \frac{\sqrt[3]{x} \ln(|\ln x|)}{\sqrt[3]{(2x+3)\ln x}},$$
which is the only thing that makes sense to me. Suppose $0<x<1/e.$ Then the absolute value of the expression in $(1)$ is
$$\frac{|x|^{1/3} \ln(|\ln x|)}{(2x+3)^{1/3}|\ln x|^{1/3}}= \frac{|x|^{1/3}}{(2x+3)^{1/3}}\frac{\ln(|\ln x|)}{|\ln x|^{1/3}} .$$
On the right, the first fraction $\to 0/3^{1/3}=0.$ Because $(\ln u)/u^{1/3} \to 0$ as $u\to \infty,$ the second fraction $\to 0.$ The desired limit is thus $0.$
