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Question Number 138767 by bramlexs22 last updated on 18/Apr/21
lim_(x→0)  ((tan (πcos^2 x))/(sin (2πsin^2 x))) =?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi\mathrm{cos}\:^{\mathrm{2}} {x}\right)}{\mathrm{sin}\:\left(\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} {x}\right)}\:=? \\ $$
Answered by EDWIN88 last updated on 18/Apr/21
 lim_(x→0)  ((tan (πcos^2 x))/(sin (2πsin^2 x)))  = lim_(x→0)  ((2πsin^2 x)/(sin (2πsin^2 x))).lim_(x→0)  ((tan (π−πsin^2 x))/(2πsin^2 x))  = 1 .lim_(x→0) ((−tan (πsin^2 x))/(2πsin^2 x)) =−(1/2)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{sin}\:\left(\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{sin}\:\left(\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi−\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\mathrm{1}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{tan}\:\left(\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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