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