# Find a limit $\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$ [duplicate]

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...

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## marked as duplicate by some1.new4u, egreg, Daniel Robert-Nicoud, mrf, Hagen von EitzenDec 2 '13 at 22:20

which part of the solution do you not understand ? – AbKDs Dec 2 '13 at 19:19
Well, there were only hints and neither I understood and I tried, believe me. – L_McClain Dec 2 '13 at 19:23
I just need to know how to get that minus infinity to plus infinity, so what you get when you convert this limit so it goes to plus infinity and with what I need to multiply or divide that limit in order to solve it to the end. That's all. – L_McClain Dec 2 '13 at 19:26
I think answer is 0 because وDenominator is greater than numerator – Khosrotash Dec 2 '13 at 19:27

$$\left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$

$$\left(\frac{4^{x}*16- 2\cdot3^{-x}}{4^{-x}+2\cdot3\cdot3^{x}}\right)$$

$$\left(\frac{4^{x}*16- 2\cdot\frac{1}{3^x}}{\frac{1}{4^x}+2\cdot3\cdot3^{x}}\right)$$

$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{3^x}\right)\cdot\left(\frac{4^x}{1+6\cdot3^x\cdot4^x}\right)$$

$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{1+6\cdot3^x\cdot4^x}\right)\cdot\left(\frac{4^x}{3^x}\right)$$

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 $0$

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How can be $4^{x}/8$ the same as $4^{x+2}$. I thought $4^{x+2}=4^{x}*16$. – L_McClain Dec 2 '13 at 20:04
Oh sorry . I will correct it ! – AbKDs Dec 2 '13 at 20:05

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.

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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 $$\frac{4^{x+2}-2\cdot 3^{-x}}{4^{-x}+2\cdot3^{x+1}}=\frac{3^{-x}}{4^{-x}}\frac{(3\cdot4)^x4^{2}-2}{1+2\cdot 3(3\cdot 4)^x}=\left(\frac{4}{3}\right)^{x}\frac{-2+8\cdot12^x}{1+6\cdot 12^x}\to0 \qquad\text{for}\; x\to-\infty$$

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