Question Number 105584 by bemath last updated on 30/Jul/20
![lim_(x→0) [csc^2 (2x)−(1/(4x^2 )) ]?](https://www.tinkutara.com/question/Q105584.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\mathrm{csc}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\:\right]? \\ $$
Answered by bobhans last updated on 30/Jul/20
![lim_(x→0) [(1/(sin^2 (2x)))−(1/(4x^2 )) ]= lim_(x→0) (((2x+sin 2x)(2x−sin 2x))/(4x^2 sin^2 (2x))) =lim_(x→0) (((2x+(2x−(((2x)^3 )/6)))(2x−(2x−(((2x)^3 )/6))))/((4x^2 )(4x^2 ))) = lim_(x→0) (((4x−((4x^3 )/3))(((4x^3 )/3)))/(16x^4 )) = lim_(x→0) ((4x(1−(1/3)x^2 ))/x) × lim_(x→0) ((4x^3 )/(3.16x^3 )) = 4×(1/(12)) = (1/3).](https://www.tinkutara.com/question/Q105588.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)}−\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\:\right]=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{x}+\mathrm{sin}\:\mathrm{2}{x}\right)\left(\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}\right)}{\mathrm{4}{x}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}{x}+\left(\mathrm{2}{x}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{6}}\right)\right)\left(\mathrm{2}{x}−\left(\mathrm{2}{x}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{6}}\right)\right)}{\left(\mathrm{4}{x}^{\mathrm{2}} \right)\left(\mathrm{4}{x}^{\mathrm{2}} \right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{4}{x}−\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}\right)\left(\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}\right)}{\mathrm{16}{x}^{\mathrm{4}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} \right)}{{x}} \\ $$$$×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}.\mathrm{16}{x}^{\mathrm{3}} }\:=\:\mathrm{4}×\frac{\mathrm{1}}{\mathrm{12}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}.\: \\ $$