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?
 A: 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?
A: 
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})}
{\log_2( 1+x^{-2} + x^{-3})}$$




$$\log_2( 1+x^{-2} + x^{-3}) = {\ln (1+x^{-2} + x^{-3})\over \ln 2}$$
    let $t = x^{-2} + x^{-3}$
    $${\ln (1+t) \over t\ln 2}  \overbrace{\large{=}} ^{ t \to 0} {1 \over \ln 2} \tag {1}$$.



Using (1) in our question,
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})}
{\log_2( 1+x^{-2} + x^{-3})} = (\ln 2)\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})}
{x^{-2} + x^{-3}}\tag{1*}$$



$${3^{1/x} - {{1\over 5^{1/x}}}\over (x^{-1})} = {15^{1/x} - 1\over5^{1/x}(x^{-1})} \overbrace{
\Large{=}}^{x\to \infty}  \ln(15) \tag{2}$$



using (2) in (1*), 
$$(\ln 2)\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})}
{x^{-2} + x^{-3}} = (\ln2)(\ln15)\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}}{x^{-1} + x^{-2}}\tag{2*}$$



let t = 1/x
$${\left(\sqrt{1-\cos^3 t}\right)^\prime\over (t + t^2)^\prime} = {3\sin t \times \cos^2 t\over (1 + 2t)\times 2 \times \sqrt{1-\cos^3 t}} = {3\sin t \times \cos^2 t\over (1 + 2t)\times 2 \times \sqrt{2}\sin(t/2)\sqrt{1+\cos(t)+\cos^2(t)}} = {3(\cos t/2) \times \cos^2 x\over (1 + 2t)\times  \sqrt{2}\sqrt{1+\cos(t)+\cos^2(t)}} \overbrace{\Large{=}}^{t \to 0}  {3 \over \sqrt 2}$$
    $$\tag{3}$$



using (3) in  (2*), 
$$(\ln2)(\ln15)\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}}{x^{-1} + x^{-2}} =  {3 \over \sqrt 2}\ln 2\ln 15 $$
A: $$\lim_{t\to 0}\frac{\sqrt{1-\cos ^3\left(t\right)}\cdot \left(3^t-5^{-t}\right)}{\log _2\left(\:1+t^2\:+\:t^3\right)}$$
Using Taylor approximation:
$$= \lim _{t\to 0}\left(\frac{\left(\sqrt{\frac{3}{2}}t+o(t)\right)\cdot \left(ln\left(15\right)t+o(t)\right)}{\frac{1}{\ln \left(2\right)}t^2+o(t^2)}\right)  = \color{red}{\sqrt{\frac{3}{2}}\ln \left(2\right)\ln \left(15\right)}$$
