Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
which part of the solution do you not understand ? –  trafalgar_law 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
1  
I think answer is 0 because وDenominator is greater than numerator –  darya khosrotash Dec 2 '13 at 19:27

3 Answers 3

up vote 0 down vote accepted

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

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

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

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

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

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 $0$

share|improve this answer
    
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 ! –  trafalgar_law 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.

share|improve this answer

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

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.