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Question Number 169182 by mathlove last updated on 25/Apr/22
lim_(x→π) ((πcos2x−x)/(1+cosx))=?
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\pi{cos}\mathrm{2}{x}−{x}}{\mathrm{1}+{cosx}}=? \\ $$
Commented by infinityaction last updated on 25/Apr/22
use l hospital rule p = lim_(x→π) ((−2πsin 2x − 1)/(−sin x)) p = lim_(x→π) ((2πsin 2x+ 1)/(sin x)) p = ((2π×0+1)/0) = (1/0) p = ∞
$${use}\:{l}\:{hospital}\:{rule} \\ $$$$\:\:\:\:{p}\:\:\:=\:\:\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{−\mathrm{2}\pi\mathrm{sin}\:\mathrm{2}{x}\:−\:\mathrm{1}}{−\mathrm{sin}\:{x}} \\ $$$$\:\:\:\:\:{p}\:\:\:=\:\:\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{2}\pi\mathrm{sin}\:\mathrm{2}{x}+\:\mathrm{1}}{\mathrm{sin}\:{x}} \\ $$$$\:\:\:\:\:\:{p}\:\:\:\:=\:\:\:\frac{\mathrm{2}\pi×\mathrm{0}+\mathrm{1}}{\mathrm{0}}\:=\:\frac{\mathrm{1}}{\mathrm{0}} \\ $$$$\:\:\:\:\:\:{p}\:\:\:\:\:=\:\:\:\:\infty \\ $$$$\:\:\: \\ $$
Answered by qaz last updated on 25/Apr/22
lim_(x→π) ((πcos 2x−x)/(1+cos x))=lim_(x→0) ((πcos 2x−x−π)/(1−cos x))=lim_(x→0) ((π(1−2x^2 )−x−π)/((1/2)x^2 ))=±∞
$$\underset{\mathrm{x}\rightarrow\pi} {\mathrm{lim}}\frac{\pi\mathrm{cos}\:\mathrm{2x}−\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\pi\mathrm{cos}\:\mathrm{2x}−\mathrm{x}−\pi}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\pi\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)−\mathrm{x}−\pi}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} }=\pm\infty \\ $$
Commented by JDamian last updated on 25/Apr/22
wrong
Commented by qaz last updated on 25/Apr/22
why would you default to ′ x→π^+ ′ ? why not ′ x→π^− ′ ?
$$\mathrm{why}\:\mathrm{would}\:\:\mathrm{you}\:\mathrm{default}\:\mathrm{to}\:'\:\mathrm{x}\rightarrow\pi^{+} \:'\:?\:\:\mathrm{why}\:\mathrm{not}\:'\:\mathrm{x}\rightarrow\pi^{−} \:'\:? \\ $$
Answered by Mathspace last updated on 25/Apr/22
changement x=π+t?give f(x)=((πcos(2t)−t−π)/(1−cost)) (t→0) ∼((π(1−2t^2 )−t−π)/(t^2 /2))=((−2πt^2 −t)/(t^2 /2)) =2(−2π−(1/t))=−4π−(2/t)→∞ lim f(x)=∞
$${changement}\:{x}=\pi+{t}?{give} \\ $$$${f}\left({x}\right)=\frac{\pi{cos}\left(\mathrm{2}{t}\right)−{t}−\pi}{\mathrm{1}−{cost}}\:\:\left({t}\rightarrow\mathrm{0}\right) \\ $$$$\sim\frac{\pi\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right)−{t}−\pi}{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}}=\frac{−\mathrm{2}\pi{t}^{\mathrm{2}} −{t}}{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$=\mathrm{2}\left(−\mathrm{2}\pi−\frac{\mathrm{1}}{{t}}\right)=−\mathrm{4}\pi−\frac{\mathrm{2}}{{t}}\rightarrow\infty \\ $$$${lim}\:{f}\left({x}\right)=\infty \\ $$

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