Limit with radicals, $\cos$, $\ln$ and powers $\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(x-1)^{\frac{1}{x}}-\ln{x^{\frac{1}{x}}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(\frac{x-1}{x})^{\frac{1}{x}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}\Big)}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}=0$
My answer is rather imprecise:$$\underset{x\rightarrow +\infty}\lim{\cos{\frac{1}{x^3}}}=1\implies\underset{x\rightarrow +\infty}\lim{\sqrt[6]{1-\cos{\frac{1}{x^3}}}}=0$$
$$\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}}=0?$$
I am aware of the mistake I have made by writting $0$ for an undefined term.
$$\underset{x\rightarrow +\infty}\lim{\Big(1-\frac{1}{x^3}\Big)}=1\implies \ln{\Big(1-\frac{1}{x}\Big)}<0\implies\underset{x\rightarrow +\infty}\lim{\Bigg(\frac{1}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}\Bigg)=-\infty}$$
The limit of the denumerator is $0$ $\&$ the limit of the whole expression is $0$.
How can I prove this concisely? 
 A: Your conclusion is wrong since we find an indeterminate form $\frac 0 0$ and we can't conclude that the limit is zero.
Indeed we have that by standard limits
$$\sqrt[6]{1-\cos{\frac{1}{x^3}}} = \frac1{x\sqrt[6]2}+O\left(\frac1{x^2}\right)$$
$$2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}} = \frac1x\log \frac32+O\left(\frac1{x^2}\right)$$
therefore
$${\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(x-1)^{\frac{1}{x}}-\ln{x^{\frac{1}{x}}}}}= \frac{\frac1{x^2}\left(\frac1{\sqrt[6]2}+O\left(\frac1{x}\right)\right)\left( \log \frac32+O\left(\frac1{x}\right)\right)}{{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}=$$
$$= \frac{\left(\frac1{\sqrt[6]2}+O\left(\frac1{x}\right)\right)\left( \log \frac32+O\left(\frac1{x}\right)\right)}{\frac{\ln{\Big(1-\frac{1}{x}\Big)}}{\frac1x}}\to -\frac1{\sqrt[6]2}\log \frac32$$
A: Let's rewrite the thing to $$L=\lim\limits_{x\to+0}\frac{\left( 1-\cos x^3\right)^\frac16\left(2^{-x}-3^{-x}\right)}{x\ln(1-x)}$$
Each of the things to first order will suffice, so let's start:
$$\lim\limits_{x\to0}\frac{e^x-1}{x}=1\Rightarrow
\lim\limits_{x\to0}\frac{a^x-1}{x}=\ln a,$$
$$1-\cos(x^3)=2sin^2\left(\frac{x^3}{2}\right), 
\lim\limits_{x\to0}\frac{\sin(x)}{x}=1\Rightarrow
\lim\limits_{x\to0}\frac{2sin^2\left(\frac{x^3}{2}\right)}{\left(\frac{x^3}{2}\right)^2}=2,$$
$$\lim\limits_{x\to0}\frac{\ln(1+x)}{x}=1$$
Should I do the rest? )
A: Examine the expression you have,
$$\underset{x\rightarrow \infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}\Big)}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}$$
Note that in the limit $x\rightarrow \infty$,
$$ \sqrt[6]{1-\cos\frac 1{x^3}} \rightarrow
\sqrt[6]{1-(1-\frac12 \frac 1{x^6})}
=\frac1{\sqrt[6]2}\frac1x$$
$$\frac{3^{\frac1x}-2^{\frac1x}}{6^{\frac1x}}
\rightarrow (1+\ln3 \frac1x)-( 1+\ln2 \frac1x)= \ln\frac32\frac1x$$
$$\frac1x \ln(1-\frac1x) \rightarrow - \frac1{x^2}$$
Thus, the limiting value is
$$\lim_{x\rightarrow \infty}\frac {\left( \frac1{\sqrt[6]2}\frac1x\right) \left(  \ln\frac32\frac1x\right)}{-\frac1{x^2}}
=- \frac1{\sqrt[6]2}\ln\frac32$$
