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Question-114075




Question Number 114075 by bemath last updated on 17/Sep/20
Answered by john santu last updated on 17/Sep/20
setting (1/x)=b ∧b→0  lim_(b→0)  ((1/(sin b)) − ((cos b)/(sin b))) = lim_(b→0) (((2sin^2 ((1/2)b))/(sin b)))    = 0
$${setting}\:\frac{\mathrm{1}}{{x}}={b}\:\wedge{b}\rightarrow\mathrm{0} \\ $$$$\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:{b}}\:−\:\frac{\mathrm{cos}\:{b}}{\mathrm{sin}\:{b}}\right)\:=\:\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{b}\right)}{\mathrm{sin}\:{b}}\right) \\ $$$$\:\:=\:\mathrm{0} \\ $$
Answered by Olaf last updated on 17/Sep/20
= lim_(x→∞) ((1/(((1/x))))−((1−(1/2)((1/x))^2 )/(1/x)))  = lim_(x→∞) (x−x+(1/(2x))) = 0
$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{{x}}\right)}−\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{{x}}}\right) \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}−{x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$

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