Question Number 63579 by Tawa1 last updated on 05/Jul/19
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{2}^{\mathrm{sec}\left(\mathrm{x}\right)} \:−\:\mathrm{2}^{\mathrm{cos}\left(\mathrm{x}\right)} }{\mathrm{x}^{\mathrm{2}} } \\ $$
Commented by Prithwish sen last updated on 05/Jul/19
![∵ form (0/(0 )) applying L′Hopital lim_(x→0) ((secx tanx 2^(secx) ln2 + sinx 2^(cosx) ln2)/(2x)) = (1/2) [ lim_(x→0) (((tanx)/x) ) . secx 2^(secx) ln2 + lim_(x→0) (((sinx)/x)).2^(cosx) ln2 ] =2ln2 please check.](https://www.tinkutara.com/question/Q63582.png)
$$\because\:\mathrm{form}\:\frac{\mathrm{0}}{\mathrm{0}\:}\:\mathrm{applying}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\frac{\mathrm{secx}\:\mathrm{tanx}\:\mathrm{2}^{\mathrm{secx}} \mathrm{ln2}\:+\:\mathrm{sinx}\:\mathrm{2}^{\mathrm{cosx}} \mathrm{ln2}}{\mathrm{2x}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\left(\frac{\mathrm{tanx}}{\mathrm{x}}\:\right)\:.\:\mathrm{secx}\:\mathrm{2}^{\mathrm{secx}} \mathrm{ln2}\:+\:\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right).\mathrm{2}^{\mathrm{cosx}} \mathrm{ln2}\:\right] \\ $$$$=\mathrm{2ln2}\:\mathrm{please}\:\mathrm{check}. \\ $$
Commented by mathmax by abdo last updated on 05/Jul/19
$${let}\:{use}\:{hospital}\:{theorem}\:\:{let}\:{A}\left({x}\right)=\frac{\mathrm{2}^{\frac{\mathrm{1}}{{cosx}}} −\mathrm{2}^{{cosx}} }{{x}^{\mathrm{2}} } \\ $$$${u}\left({x}\right)=\mathrm{2}^{\frac{\mathrm{1}}{{cosx}}} −\mathrm{2}^{{cosx}} \:\:\:{and}\:{v}\left({x}\right)={x}^{\mathrm{2}} \:\:\:{we}\:{have}\:\:{lim}_{{x}\rightarrow\mathrm{0}} {u}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{0}} {v}\left({x}\right)=\mathrm{0} \\ $$$${u}\left({x}\right)={e}^{\frac{{ln}\left(\mathrm{2}\right)}{{cosx}}} \:−{e}^{{ln}\left(\mathrm{2}\right){cosx}} \:\Rightarrow{u}^{'} \left({x}\right)\:=\left(\frac{{ln}\left(\mathrm{2}\right)}{{cosx}}\right)^{'} \:\:{e}^{\frac{{ln}\left(\mathrm{2}\right)}{{cosx}}} \:+{ln}\left(\mathrm{2}\right){sinx}\:{e}^{{ln}\left(\mathrm{2}\right){cosx}} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right){sinx}}{{cos}^{\mathrm{2}} {x}}\:{e}^{\frac{{ln}\left(\mathrm{2}\right)}{{cosx}}} \:+\:{ln}\left(\mathrm{2}\right){sinx}\:{e}^{{ln}\left(\mathrm{2}\right)\:{cosx}} \:={ln}\left(\mathrm{2}\right){sinx}\left\{\:\frac{{e}^{\frac{{ln}\left(\mathrm{2}\right)}{{cosx}}} }{{cos}^{\mathrm{2}} {x}}\:+{e}^{{ln}\left(\mathrm{2}\right){cosx}} \right\}\:\Rightarrow \\ $$$${u}^{\left(\mathrm{2}\right)} \left({x}\right)={ln}\left(\mathrm{2}\right){cosx}\left\{\:\frac{{e}^{\frac{{ln}\left(\mathrm{2}\right)}{{cosx}}} }{{cos}^{\mathrm{2}} {x}}\:+{e}^{{ln}\left(\mathrm{2}\right){cosx}} \right\}\:+{ln}\left(\mathrm{2}\right){sinx}\:\left\{…….\right\}^{\left(\mathrm{1}\right)} \:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} {u}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\:{ln}\left(\mathrm{2}\right)\left\{\mathrm{2}\:+\mathrm{2}\right\}=\mathrm{4}{ln}\left(\mathrm{2}\right)\:\:{also}\:{we}\:{have} \\ $$$${v}^{'} \left({x}\right)=\mathrm{2}{x}\:\:{and}\:{v}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} {v}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{A}\left({x}\right)\:=\frac{\mathrm{4}{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:=\mathrm{2}{ln}\left(\mathrm{2}\right). \\ $$
Commented by Tawa1 last updated on 05/Jul/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mathmax by abdo last updated on 05/Jul/19
$${you}\:{are}\:{welcome}. \\ $$