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Given-n-arc-tan-n-lim-x-n-1-n-




Question Number 31706 by gunawan last updated on 13/Mar/18
Given θ_n = arc tan n , lim_(x→∞)  θ_(n+1) −θ_n =
$$\mathrm{Given}\:\theta_{{n}} =\:{arc}\:\mathrm{tan}\:{n}\:,\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\theta_{{n}+\mathrm{1}} −\theta_{{n}} = \\ $$
Commented by abdo imad last updated on 18/Mar/18
if the question is find lim_(n→∞) θ_(n+1)  −θ_n  we have  lim θ_(n+1)  −θ_n =lim_(n→∞)  arctan(n+1) −arctann  (π/2) −(π/2)  =0
$${if}\:{the}\:{question}\:{is}\:{find}\:{lim}_{{n}\rightarrow\infty} \theta_{{n}+\mathrm{1}} \:−\theta_{{n}} \:{we}\:{have} \\ $$$${lim}\:\theta_{{n}+\mathrm{1}} \:−\theta_{{n}} ={lim}_{{n}\rightarrow\infty} \:{arctan}\left({n}+\mathrm{1}\right)\:−{arctann} \\ $$$$\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{2}}\:\:=\mathrm{0} \\ $$

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