$\lim_{x\to \infty}{\frac{4^{x+1} + 3^{x-2} + 2^{x+1}}{4^{x-1} + 3^{x+1} + 2^{x-1}}}$

Question

I found this limit problem very hard with my current high school knowledge. The answer is $16$, but I don't know how to find it. Can someone please help me?

$$\lim_{x\to \infty}{\frac{4^{x+1} + 3^{x-2} + 2^{x+1}}{4^{x-1} + 3^{x+1} + 2^{x-1}}}$$

Answer 1

$$\lim_{x\to \infty}{\frac{4^{x+1} + 3^{x-2} + 2^{x+1}}{4^{x-1} + 3^{x+1} + 2^{x-1}}}\\ =\lim_{x\rightarrow\infty}\frac{4+3^{-2}(\frac{3}{4})^x+2(\frac{1}{2})^x}{4^{-1}+3(\frac{3}{4})^x+2^{-1}(\frac{1}{2})^x}\\ =\lim_{x\rightarrow\infty}\frac{4+o(4)}{4^{-1}+o(4)}\\=16 $$and this is the answer

Answer 2

$$\lim_{x\to \infty}{\frac{4^{x+1} + 3^{x-2} + 2^{x+1}}{4^{x-1} + 3^{x+1} + 2^{x-1}}}=\lim_{x\to \infty} \frac{4^x}{4^x}{\frac{4 + \frac{3^{x-2}}{4^x} + \frac{2^{x+1}}{4^x}}{\frac14+ \frac{3^{x+1}}{4^x} + \frac{ 2^{x-1} } {4^x}}}=16$$