How to determine the limit of this sum? I know that $\lim_{x\to\infty} \dfrac{2x^5\cdot2^x}{3^x} = 0$. But what I can't figure out is how to get that answer. One of the things I tried is $\lim_{x\to\infty} 2x^5 \cdot \lim_{x\to\infty}(\dfrac{2}{3})^x$, but then you'd get $\infty \cdot 0$, and I think that is undefined. What would be a correct way to get $0$?
 A: As you can see, and this is the beauty of this site, many different ways could be proposed to approach the solution. So, let me propose another one.
Consider the function  $$f(x)=\dfrac{2x^5\cdot2^x}{3^x}$$ After some simplication work the derivative is $$f'(x)=\frac{2^{x+1}} {3^{x}} x^4 \left(5-x \log \left(\frac{3}{2}\right)\right)$$ which cancels for $x=\frac{5}{\log \left(\frac{3}{2}\right)}$. The value of the function is quite large at this point and the second derivative test confirms that it is a maximum. So, since the function is always positive, after this specific point, it continues decreasing (remember that the derivative is negative) for ever to $0^+$.
It is even easier considering $$g(x)=\log\big(f(x)\big)= \log(2)+5\log(x)+x\log \left(\frac{2}{3}\right)=\log(2)+5\log(x)-x\log \left(\frac{3}{2}\right)$$ $$g'(x)=\frac{5}{x}-\log \left(\frac{3}{2}\right)$$ Very quickly, the $\log(x)$ has vey little weight on the value of the function which almost behaves as $-x$; so, $g(x)$ decreases almost linearly when $x$ increases and goes to $-\infty$; so, $f(x)$ goes to $0^+$
A: Show that $\exists$ some $x_0$ s.t. $\forall \ x>x_0 \ x^5 < (1.4)^x$. Then use squeeze theorem.  
A: $$F=\lim_{x\to\infty}\frac{2x^5}{(3/2)^x}$$ which is of the form $\dfrac\infty\infty$
If L'Hospital's rule is allowed,
$$F=\lim_{x\to\infty}\frac{2\cdot5x^4}{(3/2)^x\ln3/2}$$ which is again of the form $\dfrac\infty\infty$
So, we can apply L'Hospital's rule again and so on
