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Find-lim-x-2-2-sin-4x-1-cos-4x-Find-lim-x-2-sin-6x-1-cos-6x-




Question Number 189757 by Shrinava last updated on 21/Mar/23
Find:     lim_(x→(𝛑/2))  (((√2) ∣ sin 4x ∣)/( (√(1 − cos 4x))))   =   ?  Find:     lim_(x→𝛑)  (((√2) ∣ sin 6x ∣)/( (√(1 − cos 6x))))   =   ?
$$\mathrm{Find}:\:\:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:\mid\:\mathrm{sin}\:\mathrm{4x}\:\mid}{\:\sqrt{\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{4x}}}\:\:\:=\:\:\:? \\ $$$$\mathrm{Find}:\:\:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\boldsymbol{\pi}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:\mid\:\mathrm{sin}\:\mathrm{6x}\:\mid}{\:\sqrt{\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{6x}}}\:\:\:=\:\:\:? \\ $$
Answered by Frix last updated on 21/Mar/23
Use t=tan 2x in the 1^(st)  and t=tan 3x in the  2^(nd)  with t→0 ⇒ answer is 2
$$\mathrm{Use}\:{t}=\mathrm{tan}\:\mathrm{2}{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{and}\:{t}=\mathrm{tan}\:\mathrm{3}{x}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{with}\:{t}\rightarrow\mathrm{0}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2} \\ $$
Answered by mehdee42 last updated on 21/Mar/23
1)lim_(x→(π/2)) ((2(√2)∣sin2xcos2x∣)/( (√2)∣sin2x∣))=lim_(x→(π/2)) 2∣cos2x∣=2  2)lim_(x→π) ((2(√2)∣sin3xcos3x∣)/( (√2)∣sin3x∣))=lim_(x→π) 2∣cos3x∣=2
$$\left.\mathrm{1}\right){li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{m}}\frac{\mathrm{2}\sqrt{\mathrm{2}}\mid{sin}\mathrm{2}{xcos}\mathrm{2}{x}\mid}{\:\sqrt{\mathrm{2}}\mid{sin}\mathrm{2}{x}\mid}={li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{m}}\mathrm{2}\mid{cos}\mathrm{2}{x}\mid=\mathrm{2} \\ $$$$\left.\mathrm{2}\right){li}\underset{{x}\rightarrow\pi} {{m}}\frac{\mathrm{2}\sqrt{\mathrm{2}}\mid{sin}\mathrm{3}{xcos}\mathrm{3}{x}\mid}{\:\sqrt{\mathrm{2}}\mid{sin}\mathrm{3}{x}\mid}={li}\underset{{x}\rightarrow\pi} {{m}}\mathrm{2}\mid{cos}\mathrm{3}{x}\mid=\mathrm{2} \\ $$$$ \\ $$
Commented by Shrinava last updated on 24/Mar/23
thank you professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$

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