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Question Number 83690 by jagoll last updated on 05/Mar/20

lim_(x→∞ ) ((tan (((πx+1)/(2x+2))))/(x+1)) = ?

$$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{\pi{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\right)}{{x}+\mathrm{1}}\:=\:? \\ $$

Commented by mathmax by abdo last updated on 05/Mar/20

let f(x)=((tan(((πx+1)/(2x+2))))/(x+1)) ⇒f(x)=((tan((π/2)×((x+(1/π))/(x+1))))/(x+1)) =((tan((π/2)×(((x+1 +(1/π)−1)/(x+1)))))/(x+1)) =((tan((π/2)(1+((1−π)/(π(x+1)))))/(x+1)) =((tan((π/2)+(((1−π))/(2(x+1)))))/(x+1)) =−(1/(tan(((1−π)/(2(x+1))))(x+1))) ⇒f(x)∼−(1/(((1−π)/(2(x+1)))×(x+1))) =((−2)/(1−π)) =(2/(π−1)) ⇒ lim_(x→+∞) f(x) =(2/(π−1))

$${let}\:{f}\left({x}\right)=\frac{{tan}\left(\frac{\pi{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\right)}{{x}+\mathrm{1}}\:\Rightarrow{f}\left({x}\right)=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}×\frac{{x}+\frac{\mathrm{1}}{\pi}}{{x}+\mathrm{1}}\right)}{{x}+\mathrm{1}} \\ $$$$=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}×\left(\frac{{x}+\mathrm{1}\:+\frac{\mathrm{1}}{\pi}−\mathrm{1}}{{x}+\mathrm{1}}\right)\right)}{{x}+\mathrm{1}}\:=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}−\pi}{\pi\left({x}+\mathrm{1}\right)}\right)\right.}{{x}+\mathrm{1}} \\ $$$$=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}+\frac{\left(\mathrm{1}−\pi\right)}{\mathrm{2}\left({x}+\mathrm{1}\right)}\right)}{{x}+\mathrm{1}}\:=−\frac{\mathrm{1}}{{tan}\left(\frac{\mathrm{1}−\pi}{\mathrm{2}\left({x}+\mathrm{1}\right)}\right)\left({x}+\mathrm{1}\right)} \\ $$$$\Rightarrow{f}\left({x}\right)\sim−\frac{\mathrm{1}}{\frac{\mathrm{1}−\pi}{\mathrm{2}\left({x}+\mathrm{1}\right)}×\left({x}+\mathrm{1}\right)}\:=\frac{−\mathrm{2}}{\mathrm{1}−\pi}\:=\frac{\mathrm{2}}{\pi−\mathrm{1}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:=\frac{\mathrm{2}}{\pi−\mathrm{1}} \\ $$

Answered by john santu last updated on 05/Mar/20

lim_(x→∞) (1/(x+1)) cot ((π/2)−((πx+1)/(2x+2))) = lim_(x→∞ ) ((cot (((2(π−1))/(4(x+1)))))/(x+1)) = lim_(x→∞) (1/((x+1)tan (((π−1)/(2(x+1)))))) × (((((π−1)/(2(x+1)))))/((((π−1)/(2(x+1)))))) = lim_(x→∞) ((2(x+1))/((π−1)(x+1))) = (2/(π−1)) ⇐ the answer

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\mathrm{cot}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\pi{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\right)\:= \\ $$$$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\left(\frac{\mathrm{2}\left(\pi−\mathrm{1}\right)}{\mathrm{4}\left({x}+\mathrm{1}\right)}\right)}{{x}+\mathrm{1}}\:=\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\mathrm{tan}\:\left(\frac{\pi−\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\right)}\:×\:\frac{\left(\frac{\pi−\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\right)}{\left(\frac{\pi−\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\right)}\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}\left({x}+\mathrm{1}\right)}{\left(\pi−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}\:=\:\frac{\mathrm{2}}{\pi−\mathrm{1}}\:\Leftarrow\:\mathrm{the}\:\mathrm{answer} \\ $$

Commented by jagoll last updated on 05/Mar/20

thank you mister

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

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