Question Number 111080 by bemath last updated on 02/Sep/20
$$\:\:\sqrt{\mathrm{bemath}} \\ $$$$\left(\mathrm{1}\right)\underset{\mathrm{k}=\mathrm{50}} {\overset{\mathrm{100}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{151}−\mathrm{k}\right)}\:? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{without}\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{and}\:\mathrm{series}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{xcos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\mathrm{x}} \\ $$
Answered by john santu last updated on 02/Sep/20
![(2) lim_(x→0) ((xcos x−sin x+x−x)/(x^2 sin x)) = lim_(x→0) ((x(cos x−1))/(x^2 sin x))+lim_(x→0) ((x−sin x)/(x^2 sin x)) = the first term L_1 = lim_(x→0) ((x(cos x−1))/(x^2 sin x))=lim_(x→0) ((cos x−1)/(xsin x)) L_1 =lim_(x→0) ((−2sin^2 ((x/2)))/(xsin x))=−2×(1/4)=−(1/2) the second term L_2 = lim_(x→0) ((x−sin x)/(x^2 sin x)) = lim_(x→0) ((x−sin x)/x^3 ) .(x/(sin x)) [ note that lim_(x→0) (x/(sin x)) = 1 ] L_2 =lim_(x→0) ((x−sin x)/x^3 ) = (1/6) [ we can find it by letting x = 2t ] Hence , we conclude that lim_(x→0) ((xcos x−sin x)/(x^2 sin x)) = −(1/2)+(1/6)=−(1/3)](https://www.tinkutara.com/question/Q111081.png)
$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\mathrm{cos}\:{x}−\mathrm{sin}\:{x}+{x}−{x}}{{x}^{\mathrm{2}} \mathrm{sin}\:{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)}{{x}^{\mathrm{2}} \mathrm{sin}\:{x}}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \mathrm{sin}\:{x}}\:= \\ $$$${the}\:{first}\:{term} \\ $$$${L}_{\mathrm{1}} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)}{{x}^{\mathrm{2}} \mathrm{sin}\:{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{1}}{{x}\mathrm{sin}\:{x}} \\ $$$${L}_{\mathrm{1}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{x}\mathrm{sin}\:{x}}=−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${the}\:{second}\:{term}\: \\ $$$${L}_{\mathrm{2}} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \mathrm{sin}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\:.\frac{{x}}{\mathrm{sin}\:{x}} \\ $$$$\left[\:{note}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{sin}\:{x}}\:=\:\mathrm{1}\:\right] \\ $$$${L}_{\mathrm{2}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\left[\:{we}\:{can}\:{find}\:{it}\:{by}\:{letting}\:{x}\:=\:\mathrm{2}{t}\:\right] \\ $$$${Hence}\:,\:{we}\:{conclude}\:{that}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \mathrm{sin}\:{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by bemath last updated on 02/Sep/20
$$\mathrm{greatt}…. \\ $$