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Question Number 90077 by manuel__ last updated on 21/Apr/20
lim_(x→∞) (sin (x+(1/x))−sin(x))=?
$${lim}_{{x}\rightarrow\infty} \left(\mathrm{sin}\:\left({x}+\frac{\mathrm{1}}{{x}}\right)−{sin}\left({x}\right)\right)=? \\ $$
Commented by jagoll last updated on 21/Apr/20
sin (x+(1/x))−sin (x) = 2cos (((2x+(1/x))/2)) sin ((1/(2x))) = 2cos (x+(1/(2x))) sin ((1/(2x))) [ let (1/(2x)) = t , t →0 ] lim_(t→0) 2cos (t+(1/(2t))) sin t = 0
$$\mathrm{sin}\:\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{sin}\:\left(\mathrm{x}\right)\:=\: \\ $$$$\mathrm{2cos}\:\left(\frac{\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{x}}}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2x}}\right)\:= \\ $$$$\mathrm{2cos}\:\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2x}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2x}}\right) \\ $$$$\left[\:\mathrm{let}\:\frac{\mathrm{1}}{\mathrm{2x}}\:=\:\mathrm{t}\:,\:\mathrm{t}\:\rightarrow\mathrm{0}\:\right] \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{2cos}\:\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2t}}\right)\:\mathrm{sin}\:\mathrm{t}\:=\:\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 21/Apr/20
let f(x)=sin(x+(1/x))−sinx f is continue ⇒ lim_(x→+∞) f(x) =lim_(x→+∞) (sinx−sinx)=0
$${let}\:{f}\left({x}\right)={sin}\left({x}+\frac{\mathrm{1}}{{x}}\right)−{sinx}\:\:{f}\:{is}\:{continue}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:={lim}_{{x}\rightarrow+\infty} \left({sinx}−{sinx}\right)=\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 21/Apr/20
another way f(x)=sinx cos((1/x))+cosx sin((1/x))−sinx ∼sinx(1−(1/(2x^2 ))) +cosx ×(1/x)−sinx (x→∞) ⇒ f(x)∼−((sinx)/(2x^2 )) +((cosx)/x) we have ∣((−sinx)/(2x^2 ))∣≤(1/(2x^2 ))→0 also ∣((cosx)/x)∣≤(1/(∣x∣))→0 ⇒lim_(x→∞) f(x)=0
$${another}\:{way}\:\:{f}\left({x}\right)={sinx}\:{cos}\left(\frac{\mathrm{1}}{{x}}\right)+{cosx}\:{sin}\left(\frac{\mathrm{1}}{{x}}\right)−{sinx} \\ $$$$\sim{sinx}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right)\:\:+{cosx}\:×\frac{\mathrm{1}}{{x}}−{sinx}\:\:\:\left({x}\rightarrow\infty\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\sim−\frac{{sinx}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\frac{{cosx}}{{x}}\:\:{we}\:{have}\:\mid\frac{−{sinx}}{\mathrm{2}{x}^{\mathrm{2}} }\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\rightarrow\mathrm{0}\:{also} \\ $$$$\mid\frac{{cosx}}{{x}}\mid\leqslant\frac{\mathrm{1}}{\mid{x}\mid}\rightarrow\mathrm{0}\:\Rightarrow{lim}_{{x}\rightarrow\infty} {f}\left({x}\right)=\mathrm{0} \\ $$

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