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Question Number 213241 by RoseAli last updated on 01/Nov/24
lim_(x→0) ((sin x−tan x)/x^3 )
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}−\mathrm{tan}\:{x}}{{x}^{\mathrm{3}} } \\ $$
Answered by ajfour last updated on 01/Nov/24
=−lim_(x→0) {((sin x)/x)×(((1−cos x)/(4((x/2))^2 )))×(1/(cos x))} =−lim_(x→0) {((sin x)/x)×(2/4)(((sin (x/2))/(x/2)))^2 ×(1/(cos x))} =−1×(2/4)×(1/1)=−(1/2)
$$=−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{sin}\:{x}}{{x}}×\left(\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{4}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)×\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right\} \\ $$$$=−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{sin}\:{x}}{{x}}×\frac{\mathrm{2}}{\mathrm{4}}\left(\frac{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}}{\frac{{x}}{\mathrm{2}}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right\} \\ $$$$=−\mathrm{1}×\frac{\mathrm{2}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by efronzo1 last updated on 02/Nov/24
= lim_(x→0) ((tan x(cos x−1))/x^3 ) = lim_(x→0) ((cos x−1)/x^2 ) = lim_(x→0) ((−sin^2 x)/(x^2 (cos x+1))) = −(1/2)
$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)}{\mathrm{x}^{\mathrm{3}} } \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)} \\ $$$$\:\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by MrGaster last updated on 02/Nov/24
=lim_(x→0) ((sin x−((sin x)/(cos x)))/x^3 ) lim_(x→0) ((sin x(1−(1/(cos x))))/x^3 )=lim_(x→0) ((sin x(cos x−1))/(x^3 cos x)) =lim_(x→0) ((sin x(−2 sin^2 ((x/2))))/(x^3 cos x)) =lim_(x→0) ((−2sin x sin^2 ((x/2)))/(x^3 cos x))=lim_(x→0) ((−2sin x((x/2))^2 )/(x^3 cos x)) =lim_(x→0) ((−2((x/2))^2 )/(x^2 cos x))=lim_(x→0) ((−x^2 )/(2x^2 cos x)) =lim_(x→0) ((−1)/(2 cos x))=((−1)/(2 cos 0))=((−1)/2)or−(1/2)
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}−\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}}{{x}^{\mathrm{3}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right)}{{x}^{\mathrm{3}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)}{{x}^{\mathrm{3}} \mathrm{cos}\:{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}\left(−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)}{{x}^{\mathrm{3}} \mathrm{cos}\:{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:{x}\:\mathrm{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{3}} \mathrm{cos}\:{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:{x}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{{x}^{\mathrm{3}} \mathrm{cos}\:{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \mathrm{cos}\:{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \mathrm{cos}\:{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:{x}}=\frac{−\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{0}}=\frac{−\mathrm{1}}{\mathrm{2}}\mathrm{or}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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