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Question Number 149782 by iloveisrael last updated on 07/Aug/21
Answered by john_santu last updated on 07/Aug/21
(1)lim_(x→0) ((sin 6x+2sin 6xcos 4x−(3sin 6x−4sin^3 6x))/(3sin x−(3sin x−4sin^3 x))) =lim_(x→0) ((−2sin 6x+2sin 6x cos 4x+4sin^3 6x )/(4sin^3 x)) =lim_(x→0) ((sin 6x(cos 4x−1+2sin^2 6x))/(2sin^3 x)) =lim_(x→0) ((sin 6x(2sin^2 6x−2sin^2 2x))/(2sin^3 x)) =lim_(x→0) (((sin 6x)/x))^3 −lim_(x→0) ((sin 6x)/(sin x))(((sin 2x)/(sin x)))^2 =216−6×4=216−24=192 ⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢
$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{6}{x}+\mathrm{2sin}\:\mathrm{6}{x}\mathrm{cos}\:\mathrm{4}{x}−\left(\mathrm{3sin}\:\mathrm{6}{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{6}{x}\right)}{\mathrm{3sin}\:{x}−\left(\mathrm{3sin}\:{x}−\mathrm{4sin}\:^{\mathrm{3}} {x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:\mathrm{6}{x}+\mathrm{2sin}\:\mathrm{6}{x}\:\mathrm{cos}\:\mathrm{4}{x}+\mathrm{4sin}\:^{\mathrm{3}} \mathrm{6}{x}\:}{\mathrm{4sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{6}{x}\left(\mathrm{cos}\:\mathrm{4}{x}−\mathrm{1}+\mathrm{2sin}\:^{\mathrm{2}} \mathrm{6}{x}\right)}{\mathrm{2sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{6}{x}\left(\mathrm{2sin}\:^{\mathrm{2}} \mathrm{6}{x}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2}{x}\right)}{\mathrm{2sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{6}{x}}{{x}}\right)^{\mathrm{3}} −\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{6}{x}}{\mathrm{sin}\:{x}}\left(\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:{x}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{216}−\mathrm{6}×\mathrm{4}=\mathrm{216}−\mathrm{24}=\mathrm{192} \\ $$$$\:\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown \\ $$
Answered by john_santu last updated on 08/Aug/21
(2)lim_(x→0) ((1−cos ((π/2)x+sin x))/(2x^2 )) = lim_(x→0) ((2sin^2 (((πx+2sin x)/4)))/(2x^2 )) =lim_(x→0) (((sin (((πx+2sin x)/4)))/((πx+2sin x)/4)))^2 ×lim_(x→0) ((((πx+2sin x)/4)/x))^2 =lim_(x→0) (((πx+2sin x)/(4x)))^2 = (((π+2)/4))^2 =(((π+2)^2 )/(16))
$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}+\mathrm{sin}\:{x}\right)}{\mathrm{2}{x}^{\mathrm{2}} }\: \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\pi{x}+\mathrm{2sin}\:{x}}{\mathrm{4}}\right)}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\left(\frac{\pi{x}+\mathrm{2sin}\:{x}}{\mathrm{4}}\right)}{\frac{\pi{x}+\mathrm{2sin}\:{x}}{\mathrm{4}}}\right)^{\mathrm{2}} ×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\frac{\pi{x}+\mathrm{2sin}\:{x}}{\mathrm{4}}}{{x}}\right)^{\mathrm{2}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\pi{x}+\mathrm{2sin}\:{x}}{\mathrm{4}{x}}\right)^{\mathrm{2}} =\:\left(\frac{\pi+\mathrm{2}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$=\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{16}} \\ $$

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