Question Number 125722 by Mammadli last updated on 13/Dec/20
$$\underset{\boldsymbol{{x}}\rightarrow\infty} {\boldsymbol{{lim}}}\left(\frac{\mathrm{3}\boldsymbol{{x}}+\mathrm{6}}{\mathrm{3}\boldsymbol{{x}}+\mathrm{2}}\right)^{\boldsymbol{{x}}+\mathrm{2}} =? \\ $$
Answered by liberty last updated on 13/Dec/20
![let us denote h(x)=((3x+6)/(3x+2)) ; r(x)=x+2 { ((lim_(x→∞) h(x)=lim_(x→∞) ((3x+6)/(3x+2)) = 1)),((lim_(x→∞) r(x)= lim_(x→∞) (x+2)=∞)) :} use the formula lim_(x→∞) (((3x+6)/(3x+2)))^(x+2) =e^(lim_(x→∞) r(x) [h(x)−1 ]) where h(x)−1 = ((3x+6)/(3x+2))−1=(4/(3x+2)) thus lim_(x→∞) (((3x+6)/(3x+2)))^(x+2) = e^(lim_(x→∞) (x+2)((4/(3x+2)))) = e^(4/3) = (e^4 )^(1/3) .](Q125725.png)
$${let}\:{us}\:{denote}\:{h}\left({x}\right)=\frac{\mathrm{3}{x}+\mathrm{6}}{\mathrm{3}{x}+\mathrm{2}}\:;\:{r}\left({x}\right)={x}+\mathrm{2} \\ $$$$\:\begin{cases}{\underset{{x}\rightarrow\infty} {\mathrm{lim}}{h}\left({x}\right)=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}{x}+\mathrm{6}}{\mathrm{3}{x}+\mathrm{2}}\:=\:\mathrm{1}}\\{\underset{{x}\rightarrow\infty} {\mathrm{lim}}{r}\left({x}\right)=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}+\mathrm{2}\right)=\infty}\end{cases} \\ $$$${use}\:{the}\:{formula}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{3}{x}+\mathrm{6}}{\mathrm{3}{x}+\mathrm{2}}\right)^{{x}+\mathrm{2}} ={e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}{r}\left({x}\right)\:\left[{h}\left({x}\right)−\mathrm{1}\:\right]} \\ $$$${where}\:{h}\left({x}\right)−\mathrm{1}\:=\:\frac{\mathrm{3}{x}+\mathrm{6}}{\mathrm{3}{x}+\mathrm{2}}−\mathrm{1}=\frac{\mathrm{4}}{\mathrm{3}{x}+\mathrm{2}} \\ $$$${thus}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{3}{x}+\mathrm{6}}{\mathrm{3}{x}+\mathrm{2}}\right)^{{x}+\mathrm{2}} =\:{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}+\mathrm{2}\right)\left(\frac{\mathrm{4}}{\mathrm{3}{x}+\mathrm{2}}\right)} \\ $$$$=\:{e}^{\frac{\mathrm{4}}{\mathrm{3}}} \:=\:\sqrt[{\mathrm{3}}]{{e}^{\mathrm{4}} }\:.\: \\ $$$$ \\ $$