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Question Number 195148 by horsebrand11 last updated on 25/Jul/23

determinant (((lim_(x→0) ((1+x sin x−cos x)/(sin^2 x))=?)))

$$\:\:\begin{array}{|c|}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{x}\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}=?}\\\hline\end{array} \\ $$

Answered by Erico last updated on 25/Jul/23

((1+xsinx−cosx)/(sin^2 x))=((1−cosx)/(sin^2 x))+(x/(sinx)) = ((1−cosx)/((1−cos^2 x)))+(x/(sinx)) = (1/(1+cosx))+(x/(sinx)) →_(x→0) (3/2)

$$\frac{\mathrm{1}+\mathrm{xsinx}−\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{x}}{\mathrm{sinx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}−\mathrm{cosx}}{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)}+\frac{\mathrm{x}}{\mathrm{sinx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cosx}}+\frac{\mathrm{x}}{\mathrm{sinx}}\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\rightarrow}\:\:\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by dimentri last updated on 25/Jul/23

⋐

$$\:\:\:\:\underbrace{\Subset} \\ $$

Answered by deleteduser4 last updated on 25/Jul/23

=^(hop) lim_(x→0) ((2sinx+xcosx)/(sin2x)) =^(hop) lim_(x→0) ((3cosx−xsinx)/(2cosx))= =(3/2) ■

$$\overset{{hop}} {=}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{sinx}+{xcosx}}{{sin}\mathrm{2}{x}}\:\:\overset{{hop}} {=}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}{cosx}−{xsinx}}{\mathrm{2}{cosx}}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:\blacksquare \\ $$

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