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Question Number 96977 by 175 last updated on 05/Jun/20

lim_(x→5) ((x − 5)/(2 + cot^2 (2/(x −5))))

$$\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\frac{{x}\:−\:\mathrm{5}}{\mathrm{2}\:+\:\mathrm{cot}^{\mathrm{2}} \frac{\mathrm{2}}{{x}\:−\mathrm{5}}} \\ $$

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).

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}−\mathrm{5}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{x}−\mathrm{5}}\right)}}\:\:\mathrm{changement}\:\mathrm{x}−\mathrm{5}\:=\mathrm{t}\:\mathrm{give}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{t}\right) \\ $$$$=\frac{\mathrm{t}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)}}\:=\frac{\mathrm{ttan}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)}{\mathrm{2tan}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)+\mathrm{1}}\:\:\left(\mathrm{x}\rightarrow\mathrm{5}\:\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$\mathrm{g}\left(\mathrm{t}\right)\:=\frac{\mathrm{tsin}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)}{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)+\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)}\:=\frac{\mathrm{tsin}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)} \\ $$$$\mathrm{0}\leqslant\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)\leqslant\mathrm{1}\:\:\Rightarrow\mathrm{1}\leqslant\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)\leqslant\mathrm{2}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\leqslant\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{t}}\right)}\leqslant\mathrm{1}\:\Rightarrow \\ $$$$\frac{\mid\mathrm{t}\mid}{\mathrm{2}}\leqslant\mid\mathrm{g}\left(\mathrm{t}\right)\mid\leqslant\mid\mathrm{t}\mid\:\rightarrow\mathrm{0}\:\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{t}\right)\:=\mathrm{0}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{5}} \mathrm{f}\left(\mathrm{x}\right). \\ $$

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.

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir}. \\ $$

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