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Question Number 18678 by Joel577 last updated on 27/Jul/17

lim_(x→∞) ((4^(x + 1) + 2^(x +1) − 3^(x + 1) )/(4^(x − 1) + 2^(x − 1 ) + 3^(x + 1) ))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{4}^{{x}\:+\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:+\mathrm{1}} \:−\:\mathrm{3}^{{x}\:+\:\mathrm{1}} }{\mathrm{4}^{{x}\:−\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:−\:\mathrm{1}\:} +\:\mathrm{3}^{{x}\:+\:\mathrm{1}} \:} \\ $$

Answered by 433 last updated on 27/Jul/17

((4^(x+1) +2^(x+1) −3^(x+1) )/(4^(x−1) +2^(x−1) +3^(x+1) ))=(((4^(x+1) /4^(x+1) )+(2^(x+1) /4^(x+1) )−(3^(x+1) /4^(x+1) ))/((4^(x−1) /4^(x+1) )+(2^(x−1) /4^(x+1) )+(3^(x+1) /4^(x+1) ))) =((1+((1/2))^(x+1) −((3/4))^(x+1) )/((1/(16))+((((1/2))^(x+1) )/4)+((3/4))^(x+1) ))→^(x→+∞) ((1+0−0)/((1/(16))+(0/4)+0))=16

$$ \\ $$$$ \\ $$$$\frac{\mathrm{4}^{{x}+\mathrm{1}} +\mathrm{2}^{{x}+\mathrm{1}} −\mathrm{3}^{{x}+\mathrm{1}} }{\mathrm{4}^{{x}−\mathrm{1}} +\mathrm{2}^{{x}−\mathrm{1}} +\mathrm{3}^{{x}+\mathrm{1}} }=\frac{\frac{\mathrm{4}^{{x}+\mathrm{1}} }{\mathrm{4}^{{x}+\mathrm{1}} }+\frac{\mathrm{2}^{{x}+\mathrm{1}} }{\mathrm{4}^{{x}+\mathrm{1}} }−\frac{\mathrm{3}^{{x}+\mathrm{1}} }{\mathrm{4}^{{x}+\mathrm{1}} }}{\frac{\mathrm{4}^{{x}−\mathrm{1}} }{\mathrm{4}^{{x}+\mathrm{1}} }+\frac{\mathrm{2}^{{x}−\mathrm{1}} }{\mathrm{4}^{{x}+\mathrm{1}} }+\frac{\mathrm{3}^{{x}+\mathrm{1}} }{\mathrm{4}^{{x}+\mathrm{1}} }} \\ $$$$=\frac{\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}+\mathrm{1}} −\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}+\mathrm{1}} }{\frac{\mathrm{1}}{\mathrm{16}}+\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}+\mathrm{1}} }{\mathrm{4}}+\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}+\mathrm{1}} }\overset{{x}\rightarrow+\infty} {\rightarrow}\frac{\mathrm{1}+\mathrm{0}−\mathrm{0}}{\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{0}}{\mathrm{4}}+\mathrm{0}}=\mathrm{16} \\ $$

Commented by Joel577 last updated on 28/Jul/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17

l=li_(x→∞) m((4+2((1/2))^x −3((3/4))^x )/((1/4)+(1/2)((1/2))^x +3((3/4))^x ))=(4/(1/4))=16. ■

$${l}={l}\underset{{x}\rightarrow\infty} {{i}m}\frac{\mathrm{4}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} −\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}} }{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} +\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}} }=\frac{\mathrm{4}}{\frac{\mathrm{1}}{\mathrm{4}}}=\mathrm{16}.\:\blacksquare \\ $$

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