Question Number 73948 by Hardy lanes last updated on 17/Nov/19
$${lim}\:\:\:\left(\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}{{x}}\right) \\ $$$${x}\rightarrow\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 17/Nov/19
$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{{x}}\:={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }×\mathrm{4}{x} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\left(\mathrm{4}{x}\right)×\left(\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}{x}}\right)^{\mathrm{2}} =\mathrm{0}×\mathrm{1}=\mathrm{0} \\ $$
Answered by MJS last updated on 17/Nov/19
![lim_(x→0) ((sin^2 2x)/x) =lim_(x→0) (((d/dx)[sin^2 2x])/((d/dx)[x])) =lim_(x→0) ((4sin 2x cos 2x)/1) =0](https://www.tinkutara.com/question/Q73951.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}}{{x}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left[\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}\right]}{\frac{{d}}{{dx}}\left[{x}\right]}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4sin}\:\mathrm{2}{x}\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}}\:=\mathrm{0} \\ $$
Answered by $@ty@m123 last updated on 17/Nov/19
$$\:\:\:\:=\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\left(\frac{\mathrm{sin}\:\mathrm{2}{x}}{{x}}\:×\mathrm{sin}\:\mathrm{2}{x}\right) \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\left(\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}{x}}×\mathrm{2}\:×\mathrm{sin}\:\mathrm{2}{x}\right) \\ $$$$\:\:\:=\:\mathrm{2}×\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\left(\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}{x}}\right)\:×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{sin}\:\mathrm{2}{x}\right) \\ $$$$\:=\mathrm{2}×\mathrm{1}×\mathrm{0} \\ $$$$\:=\mathrm{0} \\ $$