$\lim_{x\to\infty} \frac{5\cdot5^x+3^x-4^x}{5^x +2^x+27\cdot9^x}$ How can I solve this limit.  (Here $x$ belongs to natural numbers $\Bbb{N}$.) 
$$
\lim_{x\to\infty} \dfrac{5\cdot5^x+3^x-4^x}{5^x +2^x+27\cdot9^x}$$
My try:  I tried using L'Hospital, expansions of all terms using Taylor's series, and did work out.  I was just thinking if there is any simpler method?

Since Andre already gave one way, let's see how the squeeze theorem works here:
$$ 0\le\frac{5\cdot5^x+3^x-4^x}{5^x +2^x+27\cdot9^x}\le\frac{5\cdot5^x+3^x}{27\cdot 9^x}=\frac5{27}\left(\frac59\right)^x+\frac1{27}\left(\frac13\right)^x\xrightarrow[x\to\infty]{}0+0=0$$
