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Question Number 146705 by mathmax by abdo last updated on 15/Jul/21
find lim_(x→0)  ((sin(tan(2x)−x)+1−cos(πx^2 ))/x^2 )
$$\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{sin}\left(\mathrm{tan}\left(\mathrm{2x}\right)−\mathrm{x}\right)+\mathrm{1}−\mathrm{cos}\left(\pi\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} } \\ $$
Answered by Olaf_Thorendsen last updated on 15/Jul/21
f(x) = ((sin(tan(2x)−x)+1−cos(πx))/x^2 )  f(x) ∼_0  ((sin(2x−x)+1−(1−((π^2 x^2 )/2)))/x^2 )  f(x) ∼_0  ((x+((π^2 x^2 )/2))/x^2 )  f(x) ∼_0  (1/x) →_(x→0)  ∞
$${f}\left({x}\right)\:=\:\frac{\mathrm{sin}\left(\mathrm{tan}\left(\mathrm{2}{x}\right)−{x}\right)+\mathrm{1}−\mathrm{cos}\left(\pi{x}\right)}{{x}^{\mathrm{2}} } \\ $$$${f}\left({x}\right)\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{sin}\left(\mathrm{2}{x}−{x}\right)+\mathrm{1}−\left(\mathrm{1}−\frac{\pi^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}}\right)}{{x}^{\mathrm{2}} } \\ $$$${f}\left({x}\right)\:\underset{\mathrm{0}} {\sim}\:\frac{{x}+\frac{\pi^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}}}{{x}^{\mathrm{2}} } \\ $$$${f}\left({x}\right)\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{1}}{{x}}\:\underset{{x}\rightarrow\mathrm{0}} {\rightarrow}\:\infty \\ $$

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