Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{sin}\:\left(\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{sin}\:\left(\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi−\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\mathrm{1}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{tan}\:\left(\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com