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Question Number 149959 by Mathfinity last updated on 08/Aug/21
Commented by john_santu last updated on 08/Aug/21
Commented by john_santu last updated on 08/Aug/21
Commented by john_santu last updated on 08/Aug/21
why do you take mr iloveisrael's question?
$$ \\ $$why do you take mr iloveisrael's question?
Answered by gsk2684 last updated on 08/Aug/21
i) lim_(x→0) (((2x−(((2x)^3 )/(3!))+..)+(6x−(((6x)^3 )/(3!))+..)+(10x−(((10x)^3 )/(3!))+..)−(18x−(((18x)^3 )/(3!))+..))/(3(x−(x^3 /(3!))+..)−(3x−(((3x)^3 )/(3!))+..))) lim_(x→0) ((((−8−216−1000+5832)/6)x^3 +..)/(((−3+27)/6)x^3 +..)) =(((−8−216−1000+5832)/6)/((−3+27)/6)) =((4608)/(24))=192 or lim_(x→0) ((2 sin ((6x+2x)/2) cos((6x−2x)/2) +2 cos ((10x+18x)/2) sin ((10x−18x)/2))/(4 sin^3 x)) =lim_(x→0) ((2 sin 4x. cos 2x −2 cos 14x. sin 4x)/(4 sin^3 x)) =lim_(x→0) ((2 sin 4x( cos 2x − cos 14x))/(4 sin^3 x)) =lim_(x→0) ((2 sin 4x.2 sin ((14x+2x)/2) sin ((14x−2x)/2))/(4 sin^3 x)) =lim_(x→0) (( sin 4x. sin 8x. sin 6x)/( sin^3 x)) =4×8×6=192 ii)lim_(x→0) ((1−cos ((π/2)x+sin x))/(2x^2 )) =lim_(x→0) ((2sin^2 ((π/4)x+((sin x)/2)))/(2x^2 )) =lim_(((π/4)x+((sin x)/2))→0) ((sin^2 ((π/4)x+((sin x)/2)))/(((π/4)x+((sin x)/2))^2 ))lim_(x→0) ((((π/4)x+((sin x)/2))^2 )/x^2 ) =(1)^2 lim_(x→0) ((π/4)+(1/2)((sin x)/x))^2 =((π/4)+(1/2)(1))^2 =((π/4)+(1/2))^2
$$\left.{i}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}{x}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)+\left(\mathrm{6}{x}−\frac{\left(\mathrm{6}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)+\left(\mathrm{10}{x}−\frac{\left(\mathrm{10}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)−\left(\mathrm{18}{x}−\frac{\left(\mathrm{18}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)}{\mathrm{3}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+..\right)−\left(\mathrm{3}{x}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{−\mathrm{8}−\mathrm{216}−\mathrm{1000}+\mathrm{5832}}{\mathrm{6}}{x}^{\mathrm{3}} +..}{\frac{−\mathrm{3}+\mathrm{27}}{\mathrm{6}}{x}^{\mathrm{3}} +..} \\ $$$$=\frac{\frac{−\mathrm{8}−\mathrm{216}−\mathrm{1000}+\mathrm{5832}}{\mathrm{6}}}{\frac{−\mathrm{3}+\mathrm{27}}{\mathrm{6}}} \\ $$$$=\frac{\mathrm{4608}}{\mathrm{24}}=\mathrm{192} \\ $$$${or} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{6}{x}+\mathrm{2}{x}}{\mathrm{2}}\:\mathrm{cos}\frac{\mathrm{6}{x}−\mathrm{2}{x}}{\mathrm{2}}\:+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{10}{x}+\mathrm{18}{x}}{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{10}{x}−\mathrm{18}{x}}{\mathrm{2}}}{\mathrm{4}\:\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{4}{x}.\:\mathrm{cos}\:\mathrm{2}{x}\:−\mathrm{2}\:\mathrm{cos}\:\mathrm{14}{x}.\:\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{4}\:\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{4}{x}\left(\:\mathrm{cos}\:\mathrm{2}{x}\:−\:\mathrm{cos}\:\mathrm{14}{x}\right)}{\mathrm{4}\:\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{4}{x}.\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{14}{x}+\mathrm{2}{x}}{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{14}{x}−\mathrm{2}{x}}{\mathrm{2}}}{\mathrm{4}\:\mathrm{sin}\:^{\mathrm{3}} {x}}\:\: \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\:\mathrm{sin}\:\mathrm{4}{x}.\:\mathrm{sin}\:\mathrm{8}{x}.\:\mathrm{sin}\:\mathrm{6}{x}}{\:\mathrm{sin}\:^{\mathrm{3}} {x}}\:\: \\ $$$$=\mathrm{4}×\mathrm{8}×\mathrm{6}=\mathrm{192} \\ $$$$\left.{ii}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}+\mathrm{sin}\:{x}\right)}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\underset{\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$=\left(\mathrm{1}\right)^{\mathrm{2}} \underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\right)^{\mathrm{2}} =\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$ \\ $$
Commented by john_santu last updated on 08/Aug/21
false . lim_(x→0) ((π/4)+((sin x)/(2x)))^2 ≠ (π/4)+(1/2)
$${false}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}{x}}\right)^{\mathrm{2}} \neq\:\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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