Find the limit of the expression 
Find
  $$ \lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})}
{\log_2( 1+x^{-2} + x^{-3})}$$

Firstly I replace $x$ with $1/t$. Then $t$ tends to $0$. And then I fix numerator with formulas for limits considering $t \to 0$. 
But I don't know what to do with denominator. Any help?

Since it is the denominator term that is the source of concern, note that after the substitution $x=1/t$, we have
$$\begin{align}
\log_2(1+x^{-2}+x^{-3})&=\log_2(1+t^2+t^3)\\\\
&=\left(\frac{\log_2(1+t^2+t^3)}{t^2+t^3}\right)\,(t^2+t^3)\\\\
&=\left(\frac{\log(1+t^2+t^3)}{\log(2)\,(t^2+t^3)}\right)\,(t^2+t^3)
\end{align}$$

For the numerator, we see that
$$\begin{align}
\sqrt{1-\cos^3(1/x)}&=\sqrt{1-\cos^3(t)}\\\\
&=\sqrt{1-\cos(t)}\sqrt{1+\cos(t)+\cos^2(t)}\\\\
&=\sqrt{2\sin^2(t/2)}\sqrt{1+\cos(t)+\cos^2(t)}\\\\
&=\sqrt{2}\sin(t/2)\sqrt{1+\cos(t)+\cos^2(t)}
\end{align}$$
Additionally, we have
$3^{1/x}-5^{-1/x}=3^t\left(1-e^{-\log(15)t}\right)$$

Now, we can write
$$\begin{align}
\frac{\sqrt{1-\cos^3(1/x)}\left(3^{1/x}-5^{-1/x}\right)}{\log_2(1+x^{-2}+x^{-3})}&=\frac{\sqrt{1-\cos^3(t)}\left(3^{t}-5^{-t}\right)}{\log_2(1+t^{2}+t^{3})}\\\\
&=\sqrt{2}\sqrt{1+\cos(t)+\cos^2(t)}\frac{\sin(t/2)\left(3^{t}-5^{-t}\right)}{\log_2(1+t^{2}+t^{3})}\\\\
&=\color{blue}{\sqrt{2}(3^t)\sqrt{1+\cos(t)+\cos^2(t)}}\frac{\color{red}{\sin(t/2)}\color{green}{\left(1-e^{-\log(15)t}\right)}}{\color{purple}{\log_2(1+t^{2}+t^{3})}}\\\\
&=\color{blue}{\left(\sqrt{2}(3^t)\sqrt{1+\cos(t)+\cos^2(t)}\right)}\\\\
&\times \frac{\color{red}{\sin(t/2)}}{\color{orange}{t/2}}\\\\
&\times \frac{\color{gold}{t^2+t^3}}{\color{purple}{\frac{\log(1+t^2+t^3)}{\log(2)}}}\\\\
&\times \frac{\color{green}{1-e^{-\log(15)t}}}{\log(15)t}\\\\
&\times \color{orange}{\frac{t}{2}}\times {\log(15)t}\times\frac{1}{\color{gold}{t^2+t^3}}
\end{align}$$
Can you finish now?
