Evaluating $\lim_{x\to\infty} [\log_{10}{(5\cdot6^x+2\sqrt{x^3}})-\log_{10}{(\sin{x}+4\cdot6^x+\cos{x})]}$ Given this limit:

$$\lim_{x\to\infty} [\log_{10}{(5\cdot6^x+2\sqrt{x^3}})-\log_{10}{(\sin{x}+4\cdot6^x+\cos{x})]}$$    

I simplified:
$$\lim_{x\to\infty} \log_{10}{\frac{5\cdot6^x+2\sqrt{x^3}}{\sin{x}+4\cdot6^x+\cos{x}}}$$
Now, I can't understand why I can cancel $6^x$, $2\sqrt{x^3}$, $\sin{x}$, $6^x$ and $\cos{x}$ so that the solution is $\log_{10}{\frac{5}{4}}$ as my professor told me.

$$\lim_{x\to\infty} \log_{10}{\frac{5\cdot6^x+2\sqrt{x^3}}{\sin{x}+4\cdot6^x+\cos{x}}}= \lim_{x \to \infty}\log_{10}{\frac{5+2\frac{\sqrt{x^3}}{6^x}}{\frac{\sin{x}}{6^x}+4+\frac{\cos{x}}{6^x}}}=\log_{10}\frac54$$
Edit:
\begin{align}
0 \le \lim_{x \to \infty}\frac{\sqrt{x^3}}{6^x} &\le \lim_{x \to \infty}\frac{x^2}{e^{(\log 6) x}} \le \lim_{x \to \infty}\frac{x^2}{e^x} \le \lim_{x \to \infty}\frac{x^2}{1+x+\frac{x^2}{2}+\frac{x^3}{6}} = 0
\end{align}
