Evaluate this limit $\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}$ Evaluate the limit
$$\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}.$$
I've tried and it always comes out $\frac{0}{0}$, and l'Hopital doesn't seem to help me much here, what would you do?
 A: Notice that the expressions $4^{x+3}$ and $3^{x+2}$ converge to zero more rapidly when $x\to-\infty$ than the expression $2^{x+1}$. Doing the same thing in the denominator we get
$$\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}=\lim_{x\to - \infty}\frac{-2^{x+1}}{2^{x+3}}=-\frac14$$
A: I love asymptotics. Since $4^{x}$ and $3^{x}$tends to $0$ so much faster then $2^{x}$, you can just ignore them in the limit process.
This leaves you with 
$$\lim_{x \to -\infty} \frac {-2^{x+1}}{4\cdot 2^{x+1}} = -\frac 14$$
A: We have 
$\begin{align}
\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}&=\lim_{x\to - \infty}\frac{2^{x+1}\Big(4^2(\frac{4}{2})^{x+1}-3(\frac{3}{2})^{x+1}-1\Big)}{2^{x+3}\Big((\frac{1}{2})^2(\frac{4}{2})^{x+1}+\frac{1}{2}(\frac{3}{2})^{x+1}+1\Big)}\\
&=\frac14\lim_{x\to - \infty}\frac{4^2(\frac{4}{2})^{x+1}-3(\frac{3}{2})^{x+1}-1}{(\frac{1}{2})^2(\frac{4}{2})^{x+1}+\frac{1}{2}(\frac{3}{2})^{x+1}+1}\\
&=\frac14\lim_{x\to - \infty}\frac{4^2(\frac{2}{4})^{-x-1}-3(\frac{2}{3})^{-x-1}-1}{(\frac{1}{2})^2(\frac{2}{4})^{-x-1}+\frac{1}{2}(\frac{2}{3})^{-x-1}+1}\\
&=\frac14\lim_{x\to \infty}\frac{4^2(\frac{2}{4})^{x-1}-3(\frac{2}{3})^{x-1}-1}{(\frac{1}{2})^2(\frac{2}{4})^{x-1}+\frac{1}{2}(\frac{2}{3})^{x-1}+1}\\
&=\frac{1}{4} \times (-1)=-\frac{1}{4}.
\end{align}$
where I have change the variable from $-x$ to $x$ at last.
A: Hint: Rewrite the expression by multiplying by $$\frac{4^{-x-1}}{4^{-x-1}}$$ and distributing.
A: $$\lim_{x\to-\infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}=\lim_{x\to\infty}\frac{4^{3-x}-3^{2-x}-2^{1-x}}{4^{1-x}+3^{2-x}+2^{3-x}}=\lim_{x\to\infty}\frac{4^3\cdot6^x-3^2\cdot 8^x-2\cdot12^x}{4\cdot6^x+3^2\cdot8^x+2^3\cdot12^x}$$
$$\lim_{x\to\infty}\frac{4^3\cdot(\frac{6}{12})^x-3^2\cdot(\frac{8}{12})^x-2}{4\cdot(\frac{6}{12})^x+3^2\cdot(\frac{8}{12})^x+2^3}=\frac{4^3\cdot0-3^2\cdot0-2}{4\cdot0+3^2\cdot0+2^3}=-\frac{2}{8}=-\frac{1}{4}$$
