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Question Number 186064 by TUN last updated on 31/Jan/23

lim_(x→∞)  (cos2x −cos(√(4x^2 +10)))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left({cos}\mathrm{2}{x}\:−{cos}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{10}}\right) \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 31/Jan/23

2x=t      cos 2x−cos (√(4x^2 +10)) =cost−cost(√(1+((10)/t^2 )))  =cost(1−(√(1+((10)/t^2 ))) )  x→∞      t→∞  1−(√(1+(1/t^2 ) )) →0  donc  lim_(x→∞) (cos 2x−cos (√(4x^2 +10)) )=0^

$$\mathrm{2}{x}={t}\:\:\: \\ $$$$\:\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{10}}\:=\mathrm{cos}{t}−\mathrm{cos}{t}\sqrt{\mathrm{1}+\frac{\mathrm{10}}{{t}^{\mathrm{2}} }} \\ $$$$=\mathrm{cos}{t}\left(\mathrm{1}−\sqrt{\mathrm{1}+\frac{\mathrm{10}}{{t}^{\mathrm{2}} }}\:\right) \\ $$$${x}\rightarrow\infty\:\:\:\:\:\:{t}\rightarrow\infty \\ $$$$\mathrm{1}−\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:}\:\rightarrow\mathrm{0} \\ $$$${donc}\:\:\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \left(\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{10}}\:\right)=\overset{} {\mathrm{0}} \\ $$

Answered by Frix last updated on 01/Feb/23

(√(4x^2 +10))∼2x ⇒ answer is 0

$$\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{10}}\sim\mathrm{2}{x}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$

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