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Question Number 124010 by john_santu last updated on 30/Nov/20

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

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

Answered by liberty last updated on 30/Nov/20

 (√(1−cos 3x)) = (√(1−(1−2sin^2 (((3x)/2)))) = (√(2sin^2 (((3x)/2))))   = (√2) ∣sin (((3x)/2))∣ = → { (((√2) sin (((3x)/2)) ; x≥0)),((−(√2) sin (((3x)/2)) ; x<0)) :}  (1) lim_(x→0^+ ) (((sin x+(√2) sin (((3x)/2)))/(tan (x(√2)))))= ((1+((3(√2))/2))/( (√2))) = (((√2)+3)/2)  (2) lim_(x→0^− ) (((sin x−(√2) sin (((3x)/2)))/(tan (x(√2)))))=(((√2)−3)/2)  (3) lim_(x→0) (((sin x+(√(1−cos 3x)))/(tan (x(√2))))) doesn′t exist

$$\:\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{3}{x}}\:=\:\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\right.}\:=\:\sqrt{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)} \\ $$$$\:=\:\sqrt{\mathrm{2}}\:\mid\mathrm{sin}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\mid\:=\:\rightarrow\begin{cases}{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\:;\:{x}\geqslant\mathrm{0}}\\{−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\:;\:{x}<\mathrm{0}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)}{\mathrm{tan}\:\left({x}\sqrt{\mathrm{2}}\right)}\right)=\:\frac{\mathrm{1}+\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\:=\:\frac{\sqrt{\mathrm{2}}+\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)}{\mathrm{tan}\:\left({x}\sqrt{\mathrm{2}}\right)}\right)=\frac{\sqrt{\mathrm{2}}−\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}+\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{3}{x}}}{\mathrm{tan}\:\left({x}\sqrt{\mathrm{2}}\right)}\right)\:{doesn}'{t}\:{exist} \\ $$

Commented by Mammadli last updated on 30/Nov/20

Commented by john_santu last updated on 30/Nov/20

consider a^3 +b^3  =(a+b)(a^2 +b^2 −ab)   ((2+(√5)))^(1/3)  + ((2−(√5)))^(1/3)  = (4/( (((2+(√5))^2 ))^(1/3) +(((2−(√5))^2 ))^(1/3)  +1 ))

$${consider}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} \:=\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right) \\ $$$$\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }\:+\mathrm{1}\:} \\ $$

Commented by malwan last updated on 30/Nov/20

= 1

$$=\:\mathrm{1} \\ $$

Commented by bharathkumar last updated on 30/Nov/20

sex

$${sex} \\ $$

Commented by Dwaipayan Shikari last updated on 30/Nov/20

((2+(√5)))^(1/3) +((2−(√5)))^(1/3)  =p  2+(√5)+2−(√5) +3(√(4−5))(((2+(√5)))^(1/3) +((2−(√5)))^(1/3) )=p^3   4−3p=p^3   p=1

$$\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:={p} \\ $$$$\mathrm{2}+\sqrt{\mathrm{5}}+\mathrm{2}−\sqrt{\mathrm{5}}\:+\mathrm{3}\sqrt{\mathrm{4}−\mathrm{5}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\right)={p}^{\mathrm{3}} \\ $$$$\mathrm{4}−\mathrm{3}{p}={p}^{\mathrm{3}} \\ $$$${p}=\mathrm{1} \\ $$

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