Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 151001 by EDWIN88 last updated on 17/Aug/21

lim_(x→0) ((((2+sin^2 x))^(1/3) −((1+cos 2x))^(1/3) )/(x tan x)) =?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} {x}}−\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}}{{x}\:\mathrm{tan}\:{x}}\:=?\: \\ $$$$\:\:\: \\ $$

Answered by john_santu last updated on 17/Aug/21

lim_(x→0) ((((2+sin^2 x))^(1/3) −((1+cos 2x))^(1/3) )/(x tan x)) = lim_(x→0) (((2+sin^2 x)−(1+cos 2x))/(x tan x ((((2+sin^2 x)^2 ))^(1/3) +(((2+sin^2 x)(1+cos 2x)))^(1/3) +(((1+cos 2x)^2 ))^(1/3) ))) = lim_(x→0) ((sin^2 x+1−cos 2x)/(3 (4)^(1/3) x tan x)) = lim_(x→0) ((sin^2 x+2sin^2 x)/(3(4)^(1/3) x tan x)) = (3/(3 (4)^(1/3) )) = ((2)^(1/3) /2) .

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}\:\mathrm{tan}\:\mathrm{x}}\: \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)−\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2x}\right)}{\mathrm{x}\:\mathrm{tan}\:\mathrm{x}\:\left(\sqrt[{\mathrm{3}}]{\left(\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} }\:+\sqrt[{\mathrm{3}}]{\left(\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2x}\right)}\:+\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{2}} }\:\right)} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{1}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}}\: \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{4}}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}} \\ $$$$\:=\:\frac{\mathrm{3}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\mathrm{2}}\:.\: \\ $$

Answered by mathmax by abdo last updated on 17/Aug/21

f(x)=(((2+sin^2 x)^(1/3) −(1+cos(2x))^(1/3) )/(xtanx)) sin^2 x∼x^2 ⇒(2+sin^2 x)^(1/3) ∼(2+x^2 )^(1/3) =2^(1/3) (1+(x^2 /2))^(1/3) ∼2^(1/3) (1+(x^2 /6)) cos(2x)∼1−((4x^2 )/2)=1−2x^2 ⇒(1+cos(2x))^(1/3) ∼(2−2x^2 )^(1/3) =2^(1/3) (1−x^2 )^(1/3) ∼2^(1/3) (1−(x^2 /3)) and xtanx∼x^(2 ) ⇒ f(x)∼(((^3 (√2))(1+(x^2 /6))−(^3 (√2))(1−(x^2 /3)))/x^2 ) =(((^3 (√2))((x^2 /6)+(x^2 /3)))/x^2 )=(^3 (√2))((1/6)+(1/3))=(((^3 (√2)))/2) ⇒ lim_(x→0) f(x)=(1/2)(^3 (√2))

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{2}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{xtanx}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{x}\sim\mathrm{x}^{\mathrm{2}} \:\Rightarrow\left(\mathrm{2}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \sim\left(\mathrm{2}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \sim\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$\mathrm{cos}\left(\mathrm{2x}\right)\sim\mathrm{1}−\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \:\Rightarrow\left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \sim\left(\mathrm{2}−\mathrm{2x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \sim\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}\right)\:\mathrm{and}\:\mathrm{xtanx}\sim\mathrm{x}^{\mathrm{2}\:} \:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\right)−\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}\right)}{\mathrm{x}^{\mathrm{2}} }=\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com