Question Number 137255 by Nith last updated on 31/Mar/21
Answered by Dwaipayan Shikari last updated on 31/Mar/21
$${sinx}\sim{x} \\ $$$$\mathrm{2}^{{sinx}} \sim\mathrm{2}^{{x}} \sim\mathrm{1}+{xlog}\left(\mathrm{2}\right) \\ $$$$\mathrm{3}^{{sinx}} \sim\mathrm{1}+{xlog}\left(\mathrm{3}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{{sinx}} −\mathrm{3}^{{sinx}} }{{x}}=\frac{\mathrm{1}+{xlog}\left(\mathrm{2}\right)−\mathrm{1}−{xlog}\left(\mathrm{3}\right)}{{x}}={log}\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$
Answered by JoseLuisEsponizaCasares last updated on 31/Mar/21
$${L}'{Hopital}'{s}\:{Rule} \\ $$$${if}\:\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{f}\left({x}\right)}{{g}\left({x}\right)}=\frac{\mathrm{0}}{\mathrm{0}}\:\:{then} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{f}\left({x}\right)}{{g}\left({x}\right)}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{f}\:'\left({x}\right)}{{g}\:'\left({x}\right)} \\ $$$${f}\:'\left({x}\right)=\mathrm{2}^{{sinx}} {cosxln}\mathrm{2}−\mathrm{3}^{{sinx}} {cosxln}\mathrm{3} \\ $$$${f}\:'\left(\mathrm{0}\right)=\mathrm{2}^{\mathrm{0}} \left(\mathrm{1}\right){ln}\mathrm{2}−\mathrm{3}^{\mathrm{0}} \left(\mathrm{1}\right){ln}\mathrm{3} \\ $$$$=\left(\mathrm{1}\right)\left(\mathrm{1}\right){ln}\mathrm{2}−\left(\mathrm{1}\right)\left(\mathrm{1}\right){ln}\mathrm{3}={ln}\mathrm{2}−{ln}\mathrm{3} \\ $$$$={ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$${g}\:'\left({x}\right)=\mathrm{1} \\ $$$${g}'\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${Result} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{f}\:'\left({x}\right)}{{g}\:'\left({x}\right)}=\frac{{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{1}}={ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right){s} \\ $$