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lim-x-5-x-5-2-cot-2-2-x-5-




Question Number 96977 by 175 last updated on 05/Jun/20
lim_(x→5)  ((x − 5)/(2 + cot^2 (2/(x −5))))
limx5x52+cot22x5
Answered by mathmax by abdo last updated on 06/Jun/20
let f(x) =((x−5)/(2+(1/(tan^2 ((2/(x−5)))))))  changement x−5 =t give f(x)=g(t)  =(t/(2+(1/(tan^2 ((2/t)))))) =((ttan^2 ((2/t)))/(2tan^2 ((2/t))+1))  (x→5 ⇒t→0)  g(t) =((tsin^2 ((2/t)))/(2sin^2 ((2/t))+cos^2 ((2/t)))) =((tsin^2 ((2/t)))/(1+sin^2 ((2/t))))  0≤sin^2 ((2/t))≤1  ⇒1≤1+sin^2 ((2/t))≤2 ⇒(1/2)≤(1/(1+sin^2 ((2/t))))≤1 ⇒  ((∣t∣)/2)≤∣g(t)∣≤∣t∣ →0 (t→0) ⇒lim_(x→0) g(t) =0 =lim_(x→5) f(x).
letf(x)=x52+1tan2(2x5)changementx5=tgivef(x)=g(t)=t2+1tan2(2t)=ttan2(2t)2tan2(2t)+1(x5t0)g(t)=tsin2(2t)2sin2(2t)+cos2(2t)=tsin2(2t)1+sin2(2t)0sin2(2t)111+sin2(2t)21211+sin2(2t)1t2⩽∣g(t)∣⩽∣t0(t0)limx0g(t)=0=limx5f(x).
Commented by 175 last updated on 06/Jun/20
you are a great mathematical thanks alot
Commented by abdomathmax last updated on 06/Jun/20
you are welcome sir.
youarewelcomesir.

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