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Question Number 195154 by cortano12 last updated on 25/Jul/23

lim_(x→2π) (((tan (π cos x))/(x^2 (x−5π)+4π^2 (2x−π))))=?

$$\:\:\:\:\underset{{x}\rightarrow\mathrm{2}\pi} {\mathrm{lim}}\:\left(\frac{\mathrm{tan}\:\left(\pi\:\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{2}} \left({x}−\mathrm{5}\pi\right)+\mathrm{4}\pi^{\mathrm{2}} \left(\mathrm{2}{x}−\pi\right)}\right)=? \\ $$$$ \\ $$

Answered by dimentri last updated on 25/Jul/23

x−2π=y lim_(y→0) ((tan (π cos (2π+y)))/((y+2π)^2 (y−3π)+4π^2 (2y+3π))) = lim_(y→0) ((tan (π cos y))/((y^2 +4πy+4π^2 )(y−3π)+8π^2 y+12π^3 )) = lim_(y→0) ((−tan (π(1−cos y)))/(y^3 −3πy^2 +4πy^2 −12π^2 y+4π^2 y+8π^2 y)) = lim_(y→0) ((−tan (2π sin^2 ((1/2)y)))/(y^2 (y+π))) = lim_(y→0) ((−2π sin^2 ((1/2)y))/(πy^2 )) = determinant (((−( / ))))

$$\:\: \\ $$$$\:\:{x}−\mathrm{2}\pi={y}\: \\ $$$$\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi\:\mathrm{cos}\:\left(\mathrm{2}\pi+{y}\right)\right)}{\left({y}+\mathrm{2}\pi\right)^{\mathrm{2}} \:\left({y}−\mathrm{3}\pi\right)+\mathrm{4}\pi^{\mathrm{2}} \left(\mathrm{2}{y}+\mathrm{3}\pi\right)} \\ $$$$\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi\:\mathrm{cos}\:{y}\right)}{\left({y}^{\mathrm{2}} +\mathrm{4}\pi{y}+\mathrm{4}\pi^{\mathrm{2}} \right)\left({y}−\mathrm{3}\pi\right)+\mathrm{8}\pi^{\mathrm{2}} {y}+\mathrm{12}\pi^{\mathrm{3}} } \\ $$$$\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{tan}\:\left(\pi\left(\mathrm{1}−\mathrm{cos}\:{y}\right)\right)}{{y}^{\mathrm{3}} −\mathrm{3}\pi{y}^{\mathrm{2}} +\mathrm{4}\pi{y}^{\mathrm{2}} −\mathrm{12}\pi^{\mathrm{2}} {y}+\mathrm{4}\pi^{\mathrm{2}} {y}+\mathrm{8}\pi^{\mathrm{2}} {y}} \\ $$$$\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{tan}\:\left(\mathrm{2}\pi\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)\right)}{{y}^{\mathrm{2}} \left({y}+\pi\right)} \\ $$$$\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2}\pi\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)}{\pi{y}^{\mathrm{2}} }\:=\:\begin{array}{|c|}{−\frac{ }{ }}\\\hline\end{array} \\ $$

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