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lim-x-4-x-2-3-x-2-x-4-x-1-3-x-1-




Question Number 144106 by mathdanisur last updated on 21/Jun/21
lim_(x→∞) ((4^(x-2)  + 3^x  + 2^x )/(4^(x-1)  + 3^(x+1) )) = ?
$$\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{4}^{{x}-\mathrm{2}} \:+\:\mathrm{3}^{{x}} \:+\:\mathrm{2}^{{x}} }{\mathrm{4}^{{x}-\mathrm{1}} \:+\:\mathrm{3}^{{x}+\mathrm{1}} }\:=\:? \\ $$
Commented by bobhans last updated on 22/Jun/21
 lim_(x→∞)  ((4^x (4^(−2) +((3/4))^x +((1/2))^x ))/(4^x (4^(−1) +3((3/4))^x )))  = lim_(x→∞) (((4^(−2) +((3/4))^x +((1/2))^x )/(4^(−1) +3((3/4))^x )))   = (4^(−2) /4^(−1) ) = 4^(−1) .
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{4}^{\mathrm{x}} \left(\mathrm{4}^{−\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{x}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{x}} \right)}{\mathrm{4}^{\mathrm{x}} \left(\mathrm{4}^{−\mathrm{1}} +\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{x}} \right)} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{4}^{−\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{x}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{x}} }{\mathrm{4}^{−\mathrm{1}} +\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{x}} }\right) \\ $$$$\:=\:\frac{\mathrm{4}^{−\mathrm{2}} }{\mathrm{4}^{−\mathrm{1}} }\:=\:\mathrm{4}^{−\mathrm{1}} . \\ $$
Answered by Dwaipayan Shikari last updated on 21/Jun/21
lim_(x→∞) ((4^(x−2) +3^x +2^x )/(4^(x−1) +3^(x+1) ))  =(((1/4^2 )+((3/4))^x +((2/4))^x )/((1/4)+3((3/4))^x ))=(((1/4^2 )+0+0)/((1/4)+3(0) )) =(1/4)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{4}^{{x}−\mathrm{2}} +\mathrm{3}^{{x}} +\mathrm{2}^{{x}} }{\mathrm{4}^{{x}−\mathrm{1}} +\mathrm{3}^{{x}+\mathrm{1}} } \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}} +\left(\frac{\mathrm{2}}{\mathrm{4}}\right)^{{x}} }{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}} }=\frac{\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\mathrm{0}+\mathrm{0}}{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{3}\left(\mathrm{0}\right)\:}\:=\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$
Commented by harckinwunmy last updated on 21/Jun/21
ff
$${ff} \\ $$
Commented by mathdanisur last updated on 21/Jun/21
thanks Sir
$${thanks}\:{Sir} \\ $$

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