Limit question please math help?

Question

I have to find $$\lim_{x\to\infty}\frac{2^{x+1}+3^{x+1}}{2^x-3^x}$$ .... so I thought first about separating them..and then factor what ? can you tell me just the start please?

Answer 1

$$\frac{2^{x+1}+3^{x+1}}{2^x-3^x}=3\frac{\left(\frac{2}{3}\right)^{x+1}+1}{\left(\frac{2}{3}\right)^x-1}\xrightarrow[x\to\infty]{}3\frac{0+1}{0-1}=-3$$

Answer 2

$$\lim_{x\to\infty}\frac{2^{x+1}+3^{x+1}}{2^x-3^x}=\lim_{x\to\infty}\frac{2\cdot (2/3)^x+3}{(2/3)^x-1}=-3$$ since $\lim_{x\to \infty}a^x=0,$ if $|a|<1$