Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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)=∞

$${changement}\:{x}=\pi+{t}?{give} \\ $$$${f}\left({x}\right)=\frac{\pi{cos}\left(\mathrm{2}{t}\right)−{t}−\pi}{\mathrm{1}−{cost}}\:\:\left({t}\rightarrow\mathrm{0}\right) \\ $$$$\sim\frac{\pi\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right)−{t}−\pi}{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}}=\frac{−\mathrm{2}\pi{t}^{\mathrm{2}} −{t}}{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$=\mathrm{2}\left(−\mathrm{2}\pi−\frac{\mathrm{1}}{{t}}\right)=−\mathrm{4}\pi−\frac{\mathrm{2}}{{t}}\rightarrow\infty \\ $$$${lim}\:{f}\left({x}\right)=\infty \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com