Find a limit $\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$ I am to find the limit of 
$$\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$ 
so I used:
$$\lim_{x \to -\infty} = \lim_{x \to \infty}f(-x)$$ 
but I just can't solve it to the end...
Please show me all steps, or at least most of them, so I'll know how to solve it. Thank you. 
This question was posted on: Find $\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$, and got 3 answers, but I still don't know how should I solve it, because when I try to solve it (with help of those 3 answers) I get :
$$0−12/0$$ every time and that goes to minus infinity... 

$$\left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$
$$\left(\frac{4^{x}*16- 2\cdot3^{-x}}{4^{-x}+2\cdot3\cdot3^{x}}\right)$$
$$\left(\frac{4^{x}*16- 2\cdot\frac{1}{3^x}}{\frac{1}{4^x}+2\cdot3\cdot3^{x}}\right)$$
$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{3^x}\right)\cdot\left(\frac{4^x}{1+6\cdot3^x\cdot4^x}\right)$$
$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{1+6\cdot3^x\cdot4^x}\right)\cdot\left(\frac{4^x}{3^x}\right)$$
now when you put $x \rightarrow -\infty$ the $4^x/3^x$ tends to $0$ and the limit tends to $0$
As for $$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{1+6\cdot3^x\cdot4^x}\right)$$ the limit is finite value which is ultimately gonna multipled to $0$ 
