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Question Number 62657 by mathmax by abdo last updated on 24/Jun/19
calculate lim_(x→0) ((ln(1+tan(2x))−ln(cos(3x)))/x^3 )
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{ln}\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)−{ln}\left({cos}\left(\mathrm{3}{x}\right)\right)}{{x}^{\mathrm{3}} } \\ $$
Commented by mathmax by abdo last updated on 24/Jun/19
let use hospital teorem with u(x)=ln(1+tan(2x))−ln(cos(3x)) and v(x) =x^3 we have u^′ (x) =((2(1+tan^2 (2x)))/(1+tan(2x)))+ ((3sin(3x))/(cos(3x))) =2 (((1+tan(2x))^2 −2tan(2x))/(1+tan(2x))) +3 tan(3x) =2(1+tan(2x))−4 ((tan(2x)+1−1)/(1+tan(2x))) +3tan(3x) =2(1+tan(2x))−4 +(4/(1+tan(2x))) +3 tan(3x) =−2 +2tan(2x) +3tan(3x)+(4/(1+tan(2x))) ⇒ u^((2)) (x) =4(1+tan^2 (2x)) +9(1+tan^2 (3x))+4((−2(1+tan^2 (2x)))/((1+tan(2x))^2 )) ⇒ u^((3)) (x) =4{4tan(2x)(1+tan^2 (2x))}+ 9{ 6tan(3x)(1+tan^2 (3x)} −8((4tan(2x)(1+tan^2 (2x)(1+tan(2x))^2 −(1+tan^2 (2x))2(1+tan(2x))2(1+tan^2 (2x)))/((1+tan(2x))^4 )) ⇒lim_(x→0) u^((3)) (x) =−8 ((−4)/1) =32 also we have v^′ (x)=3x^2 ⇒v^((2)) (x)=6x ⇒v^((3)) (x)=6 ⇒lim_(x→0) ((ln(1+tan(2x))−ln(cos(3x)))/x^3 ) =((32)/6) =((16)/3) .
$${let}\:{use}\:{hospital}\:{teorem}\:\:{with}\:{u}\left({x}\right)={ln}\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)−{ln}\left({cos}\left(\mathrm{3}{x}\right)\right) \\ $$$${and}\:{v}\left({x}\right)\:={x}^{\mathrm{3}} \:\:\:{we}\:{have}\: \\ $$$${u}^{'} \left({x}\right)\:=\frac{\mathrm{2}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)}{\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)}+\:\frac{\mathrm{3}{sin}\left(\mathrm{3}{x}\right)}{{cos}\left(\mathrm{3}{x}\right)} \\ $$$$=\mathrm{2}\:\:\frac{\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} −\mathrm{2}{tan}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)}\:+\mathrm{3}\:{tan}\left(\mathrm{3}{x}\right) \\ $$$$=\mathrm{2}\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)−\mathrm{4}\:\frac{{tan}\left(\mathrm{2}{x}\right)+\mathrm{1}−\mathrm{1}}{\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)}\:+\mathrm{3}{tan}\left(\mathrm{3}{x}\right) \\ $$$$=\mathrm{2}\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)−\mathrm{4}\:+\frac{\mathrm{4}}{\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)}\:+\mathrm{3}\:{tan}\left(\mathrm{3}{x}\right) \\ $$$$=−\mathrm{2}\:+\mathrm{2}{tan}\left(\mathrm{2}{x}\right)\:+\mathrm{3}{tan}\left(\mathrm{3}{x}\right)+\frac{\mathrm{4}}{\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)}\:\Rightarrow \\ $$$${u}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\mathrm{4}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)\:+\mathrm{9}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\right)+\mathrm{4}\frac{−\mathrm{2}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)}{\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${u}^{\left(\mathrm{3}\right)} \left({x}\right)\:=\mathrm{4}\left\{\mathrm{4}{tan}\left(\mathrm{2}{x}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)\right\}+\:\mathrm{9}\left\{\:\mathrm{6}{tan}\left(\mathrm{3}{x}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\right\}\right. \\ $$$$−\mathrm{8}\frac{\mathrm{4}{tan}\left(\mathrm{2}{x}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} −\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)\mathrm{2}\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)\mathrm{2}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)\right.}{\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)^{\mathrm{4}} } \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {u}^{\left(\mathrm{3}\right)} \left({x}\right)\:=−\mathrm{8}\:\frac{−\mathrm{4}}{\mathrm{1}}\:=\mathrm{32}\:\:\:{also}\:{we}\:{have}\:{v}^{'} \left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} \:\Rightarrow{v}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{6}{x}\:\Rightarrow{v}^{\left(\mathrm{3}\right)} \left({x}\right)=\mathrm{6} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)−{ln}\left({cos}\left(\mathrm{3}{x}\right)\right)}{{x}^{\mathrm{3}} }\:=\frac{\mathrm{32}}{\mathrm{6}}\:=\frac{\mathrm{16}}{\mathrm{3}}\:. \\ $$$$ \\ $$
Commented by MJS last updated on 24/Jun/19
I don′t think we can use l′Hopital a second time because u′(0)=2
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{l}'\mathrm{Hopital}\:\mathrm{a}\:\mathrm{second} \\ $$$$\mathrm{time}\:\mathrm{because}\:{u}'\left(\mathrm{0}\right)=\mathrm{2} \\ $$
Answered by MJS last updated on 24/Jun/19
lim_(x→0) ((ln (1+tan 2x) −ln cos 3x)/x^3 )= =lim_(x→0) (((d/dx)[ln (1+tan 2x) −ln cos 3x])/((d/dx)[x^3 ]))= =lim_(x→0) (((4/(1+cos 4x +sin 4x))+3tan 3x)/(3x^(2 ) ))=(2/0) ⇒ undefined
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{tan}\:\mathrm{2}{x}\right)\:−\mathrm{ln}\:\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{3}} }= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\left[\mathrm{ln}\:\left(\mathrm{1}+\mathrm{tan}\:\mathrm{2}{x}\right)\:−\mathrm{ln}\:\mathrm{cos}\:\mathrm{3}{x}\right]}{\frac{{d}}{{dx}}\left[{x}^{\mathrm{3}} \right]}= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{4}}{\mathrm{1}+\mathrm{cos}\:\mathrm{4}{x}\:+\mathrm{sin}\:\mathrm{4}{x}}+\mathrm{3tan}\:\mathrm{3}{x}}{\mathrm{3}{x}^{\mathrm{2}\:} }=\frac{\mathrm{2}}{\mathrm{0}}\:\Rightarrow\:\mathrm{undefined} \\ $$

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