Question Number 61363 by bhanukumarb2@gmail.com last updated on 01/Jun/19
Answered by tanmay last updated on 02/Jun/19
![N_r =2^(2b_n ) −2^(1+b_n ) .3^b_n +3^(2b_n ) =(2^b_n )^2 −2×2^b_n ×3^b_n +(3^b_n )^2 =(3^b_n −2^b_n )^2 b_1 =1 b_n =sin(b_(n−1) ) b_2 =sin(b_1 )=sin1 b_3 =sin(b_2 )=sin(sin1) b_4 =sin(b_3 )=sin(sin(sin1)) sin1>sin(sin1)>sin(sin(sin1)). so n→∞ b_n →0 when n→∞ b_n →0 wait pls i am trying... x=b_n lim_(x→0) (((3^x −2^x )^2 )/(cosx+1−e^x −e^(−x) ))((0/0)) =lim_(x→0) ((2(3^x −2^x )(3^x ln3−2^x ln2))/(sinx−e^x +e^(−x) )) =lim_(x→0) 2(3^x ln3−2^x ln2)×(((((3^x −1)/x))−(((2^x −1)/x)))/(((sinx)/x)−e^(−x) (((e^(2x) −1)/(2x)))×2)) =2(ln3−ln2)×(((ln3−ln2))/(1−1×1×2)) =((2[ln((3/2))]^2 )/(−1))=−2[ln((3/2))]^2 pls check is it cirrect...](Q61364.png)
$${N}_{{r}} =\mathrm{2}^{\mathrm{2}{b}_{{n}} } −\mathrm{2}^{\mathrm{1}+{b}_{{n}} } .\mathrm{3}^{{b}_{{n}} } +\mathrm{3}^{\mathrm{2}{b}_{{n}} } \\ $$$$=\left(\mathrm{2}^{{b}_{{n}} } \right)^{\mathrm{2}} −\mathrm{2}×\mathrm{2}^{{b}_{{n}} } ×\mathrm{3}^{{b}_{{n}} } +\left(\mathrm{3}^{{b}_{{n}} } \right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{3}^{{b}_{{n}} } −\mathrm{2}^{{b}_{{n}} } \right)^{\mathrm{2}} \\ $$$${b}_{\mathrm{1}} =\mathrm{1} \\ $$$${b}_{{n}} ={sin}\left({b}_{{n}−\mathrm{1}} \right) \\ $$$${b}_{\mathrm{2}} ={sin}\left({b}_{\mathrm{1}} \right)={sin}\mathrm{1} \\ $$$${b}_{\mathrm{3}} ={sin}\left({b}_{\mathrm{2}} \right)={sin}\left({sin}\mathrm{1}\right) \\ $$$${b}_{\mathrm{4}} ={sin}\left({b}_{\mathrm{3}} \right)={sin}\left({sin}\left({sin}\mathrm{1}\right)\right) \\ $$$${sin}\mathrm{1}>{sin}\left({sin}\mathrm{1}\right)>{sin}\left({sin}\left({sin}\mathrm{1}\right)\right). \\ $$$${so}\:{n}\rightarrow\infty\:{b}_{{n}} \rightarrow\mathrm{0} \\ $$$${when}\:{n}\rightarrow\infty\:{b}_{{n}} \rightarrow\mathrm{0} \\ $$$${wait}\:{pls}\:{i}\:{am}\:{trying}... \\ $$$${x}={b}_{{n}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\left(\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \right)^{\mathrm{2}} }{{cosx}+\mathrm{1}−{e}^{{x}} −{e}^{−{x}} }\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \right)\left(\mathrm{3}^{{x}} {ln}\mathrm{3}−\mathrm{2}^{{x}} {ln}\mathrm{2}\right)}{{sinx}−{e}^{{x}} +{e}^{−{x}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{2}\left(\mathrm{3}^{{x}} {ln}\mathrm{3}−\mathrm{2}^{{x}} {ln}\mathrm{2}\right)×\frac{\left(\frac{\mathrm{3}^{{x}} −\mathrm{1}}{{x}}\right)−\left(\frac{\mathrm{2}^{{x}} −\mathrm{1}}{{x}}\right)}{\frac{{sinx}}{{x}}−{e}^{−{x}} \left(\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{2}{x}}\right)×\mathrm{2}} \\ $$$$=\mathrm{2}\left({ln}\mathrm{3}−{ln}\mathrm{2}\right)×\frac{\left({ln}\mathrm{3}−{ln}\mathrm{2}\right)}{\mathrm{1}−\mathrm{1}×\mathrm{1}×\mathrm{2}} \\ $$$$=\frac{\mathrm{2}\left[{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right]^{\mathrm{2}} }{−\mathrm{1}}=−\mathrm{2}\left[{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right]^{\mathrm{2}} \\ $$$${pls}\:{check}\:{is}\:{it}\:{cirrect}... \\ $$
Commented by bhanukumarb2@gmail.com last updated on 01/Jun/19
$${right}\:{mine}\:{also}\:{same} \\ $$$$\: \\ $$
Commented by bhanukumarb2@gmail.com last updated on 01/Jun/19
$${thankuu} \\ $$