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... 
 A: Numerator: $4^{x+2}=\varepsilon_1(x) \to 0$ is always positive.
Denominator: $2 \cdot 3^{x+1} = \varepsilon_2(x) \to 0$ is also always positive and we can choose an arbitrary constant $c_1$ such that 
$$
\lim_{x \to -\infty} \frac{4^{x+2} - 2 \cdot 3^{-x}}{4^{-x} +2 \cdot 3^{x+1}}> \lim_{x \to -\infty}\frac{-2 \cdot 3^{-x}}{c_1 4^{-x}}=0
$$
Using the same logic for the upper bound, we get 
$$
\lim_{x \to -\infty} \frac{4^{x+2} - 2 \cdot 3^{-x}}{4^{-x} +2 \cdot 3^{x+1}}< \lim_{x \to -\infty}\frac{-2 c_2 \cdot 3^{-x}}{4^{-x}}=0
$$
By squeeze lemma, the limit is 0.
A: First Method
For $x\to -\infty$, $4^{x+2}\sim 0$ and $3^{x+1}\sim 0$ so
$$
\frac{4^{x+2}-2\cdot 3^{-x}}{4^{-x}+2\cdot3^{x+1}}\sim \frac{-2\cdot3^{-x}}{4^{-x}}=-2\left(\frac{4}{3}\right)^{x}\to 0\qquad\text{for}\; x\to-\infty
$$
Second Method
$$
\frac{4^{x+2}-2\cdot 3^{-x}}{4^{-x}+2\cdot3^{x+1}}=\frac{3^{-x}}{4^{-x}}\frac{(3\cdot4)^x4^{2}-2}{1+2\cdot 3(3\cdot 4)^x}=\left(\frac{4}{3}\right)^{x}\frac{-2+8\cdot12^x}{1+6\cdot 12^x}\to0 \qquad\text{for}\; x\to-\infty
$$
A: $$\left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$
$$\left(\frac{4^{x}*16- 2\cdot3^{-x}}{4^{-x}+2\cdot3\cdot3^{x}}\right)$$
$$\left(\frac{4^{x}*16- 2\cdot\frac{1}{3^x}}{\frac{1}{4^x}+2\cdot3\cdot3^{x}}\right)$$
$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{3^x}\right)\cdot\left(\frac{4^x}{1+6\cdot3^x\cdot4^x}\right)$$
$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{1+6\cdot3^x\cdot4^x}\right)\cdot\left(\frac{4^x}{3^x}\right)$$
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 $0$ 
