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lim-n-0-2-n-1-3-n-2-4-n-3-2-n-3-n-4-n-




Question Number 138249 by sahnaz last updated on 11/Apr/21
lim_(n→0) ((2^(n+1) +3^(n+2) +4^(n+3) )/(2^n +3^n +4^n ))
$$\mathrm{li}\underset{\mathrm{n}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\mathrm{2}^{\mathrm{n}+\mathrm{1}} +\mathrm{3}^{\mathrm{n}+\mathrm{2}} +\mathrm{4}^{\mathrm{n}+\mathrm{3}} }{\mathrm{2}^{\mathrm{n}} +\mathrm{3}^{\mathrm{n}} +\mathrm{4}^{\mathrm{n}} } \\ $$
Answered by mathmax by abdo last updated on 11/Apr/21
lim_(n→0)  ((2^(n+1)  +3^(n+2)  +4^(n+3) )/(2^n  +3^n  +4^n )) =((2+3^2  +4^3 )/3) =((2+9+32)/3)=((11+32)/3)=((43)/3)
$$\mathrm{lim}_{\mathrm{n}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \:+\mathrm{3}^{\mathrm{n}+\mathrm{2}} \:+\mathrm{4}^{\mathrm{n}+\mathrm{3}} }{\mathrm{2}^{\mathrm{n}} \:+\mathrm{3}^{\mathrm{n}} \:+\mathrm{4}^{\mathrm{n}} }\:=\frac{\mathrm{2}+\mathrm{3}^{\mathrm{2}} \:+\mathrm{4}^{\mathrm{3}} }{\mathrm{3}}\:=\frac{\mathrm{2}+\mathrm{9}+\mathrm{32}}{\mathrm{3}}=\frac{\mathrm{11}+\mathrm{32}}{\mathrm{3}}=\frac{\mathrm{43}}{\mathrm{3}} \\ $$

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