Question Number 116037 by Rio Michael last updated on 30/Sep/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{27}^{{x}} −\mathrm{1}}{\mathrm{9}^{{x}} −\mathrm{1}}\:=\:?? \\ $$
Answered by bemath last updated on 30/Sep/20
$${let}\:\mathrm{3}^{{x}} =\:{t}\:;\:{t}\rightarrow\mathrm{1} \\ $$$$\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{t}^{\mathrm{3}} −\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}\:=\:\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by Rio Michael last updated on 30/Sep/20
$$\mathrm{exactly}. \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{27}^{{x}} −\mathrm{1}}{\mathrm{9}^{{x}} −\mathrm{1}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{3}^{{x}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{2}{x}} +\:\left(\mathrm{3}^{{x}} \right)+\mathrm{1}\right)}{\left(\mathrm{3}^{{x}} −\mathrm{1}\right)\left(\mathrm{3}^{{x}} +\mathrm{1}\right)}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by bemath last updated on 30/Sep/20
$${yes}.. \\ $$
Answered by Dwaipayan Shikari last updated on 30/Sep/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{27}^{\mathrm{x}} −\mathrm{1}}{\mathrm{x}}.\frac{\mathrm{x}}{\mathrm{9}^{\mathrm{x}} −\mathrm{1}}=\frac{\mathrm{log}\left(\mathrm{27}\right)}{\mathrm{log}\left(\mathrm{9}\right)}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$