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Question Number 62435 by mathsolverby Abdo last updated on 21/Jun/19
calculate lim_(x→0) (((1+x)^(sinx) −1)/x^2 )
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\left(\mathrm{1}+{x}\right)^{{sinx}} −\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$
Commented by Smail last updated on 22/Jun/19
(1+x)^(sinx) =e^(ln((1+x)^(sinx) )) =e^(sinx×ln(1+x)) ln(1+x)∼x when x is near 0 sinx∼x when x is near 0 so (1+x)^(sinx) ∼e^(x×x) ∼e^x^2 e^x ∼1+x when x near 0 so (1+x)^(sinx) ∼1+x^2 lim_(x→0) (((1+x)^(sinx) −1)/x^2 )=lim_(x→0) ((1+x^2 −1)/x^2 ) =1
$$\left(\mathrm{1}+{x}\right)^{{sinx}} ={e}^{{ln}\left(\left(\mathrm{1}+{x}\right)^{{sinx}} \right)} ={e}^{{sinx}×{ln}\left(\mathrm{1}+{x}\right)} \\ $$$${ln}\left(\mathrm{1}+{x}\right)\sim{x}\:\:{when}\:{x}\:{is}\:{near}\:\mathrm{0} \\ $$$${sinx}\sim{x}\:\:{when}\:\:{x}\:{is}\:{near}\:\mathrm{0} \\ $$$${so}\:\left(\mathrm{1}+{x}\right)^{{sinx}} \sim{e}^{{x}×{x}} \sim{e}^{{x}^{\mathrm{2}} } \\ $$$${e}^{{x}} \sim\mathrm{1}+{x}\:{when}\:{x}\:{near}\:\mathrm{0} \\ $$$${so}\:\:\left(\mathrm{1}+{x}\right)^{{sinx}} \sim\mathrm{1}+{x}^{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{x}\right)^{{sinx}} −\mathrm{1}}{{x}^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$=\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 23/Jun/19
let use hospital theorem u(x) =(1+x)^(sinx) −1 ⇒u(x) =e^(sinxln(1+x)) −1 ⇒ u^′ (x) =(cosx ln(1+x) +((sinx)/(1+x)))e^(sinxln(1+x)) ⇒ u^((2)) (x) ={ −sinxln(1+x)+((cosx)/(1+x)) +(((1+x)cosx−sinx)/((1+x)^2 )))e^(sinxln(1+x)) +(cosx ln(1+x)+((sinx)/(1+x)))^2 e^(sinxln(1+x)) ⇒lim_(x→0) u^((2)) (x)= 2 v(x) =x^2 ⇒g^′ (x) =2x and v^((2)) (x) =2 =lim_(x→0) g^((2)) (x) ⇒ lim_(x→0) (((1+x)^(sinx) −1)/x^2 ) =lim_(x→0) ((u^((2)) (x))/(v^((2)) (x))) =(2/2) =1 .
$${let}\:{use}\:{hospital}\:{theorem}\:\:{u}\left({x}\right)\:=\left(\mathrm{1}+{x}\right)^{{sinx}} −\mathrm{1}\:\Rightarrow{u}\left({x}\right)\:={e}^{{sinxln}\left(\mathrm{1}+{x}\right)} −\mathrm{1}\:\Rightarrow \\ $$$${u}^{'} \left({x}\right)\:=\left({cosx}\:{ln}\left(\mathrm{1}+{x}\right)\:+\frac{{sinx}}{\mathrm{1}+{x}}\right){e}^{{sinxln}\left(\mathrm{1}+{x}\right)} \:\Rightarrow \\ $$$${u}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\left\{\:−{sinxln}\left(\mathrm{1}+{x}\right)+\frac{{cosx}}{\mathrm{1}+{x}}\:+\frac{\left(\mathrm{1}+{x}\right){cosx}−{sinx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\right){e}^{{sinxln}\left(\mathrm{1}+{x}\right)} \\ $$$$+\left({cosx}\:{ln}\left(\mathrm{1}+{x}\right)+\frac{{sinx}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} \:{e}^{{sinxln}\left(\mathrm{1}+{x}\right)} \:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{u}^{\left(\mathrm{2}\right)} \left({x}\right)=\:\mathrm{2} \\ $$$${v}\left({x}\right)\:={x}^{\mathrm{2}} \:\Rightarrow{g}^{'} \left({x}\right)\:=\mathrm{2}{x}\:{and}\:{v}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\mathrm{2}\:={lim}_{{x}\rightarrow\mathrm{0}} {g}^{\left(\mathrm{2}\right)} \left({x}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\left(\mathrm{1}+{x}\right)^{{sinx}} −\mathrm{1}}{{x}^{\mathrm{2}} }\:={lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{u}^{\left(\mathrm{2}\right)} \left({x}\right)}{{v}^{\left(\mathrm{2}\right)} \left({x}\right)}\:=\frac{\mathrm{2}}{\mathrm{2}}\:=\mathrm{1}\:. \\ $$

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