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Question Number 187270 by MikeH last updated on 15/Feb/23

lim_(x→π) ((cos x + sin(2x) +1)/(x^2 −π^2 )) = A. (1/(2π)) C. 1 B. (1/π) D. Does Not exist

$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}\:+\:\mathrm{sin}\left(\mathrm{2}{x}\right)\:+\mathrm{1}}{{x}^{\mathrm{2}} −\pi^{\mathrm{2}} }\:= \\ $$$$\mathrm{A}.\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\:\:\:\:\:\:\:\:{C}.\:\mathrm{1} \\ $$$${B}.\:\frac{\mathrm{1}}{\pi}\:\:\:\:\:\:\:\:\:{D}.\:\mathrm{Does}\:\mathrm{Not}\:\mathrm{exist} \\ $$

Answered by horsebrand11 last updated on 15/Feb/23

lim_(x→π) ((1−cos (π−x)−sin 2(π−x))/((x+π)(x−π))) = (1/(2π)) .lim_(x→0) ((1−cos x−sin 2x)/(−x)) =−(1/(2π)) [ lim_(x→0) ((1−cos x)/x) −lim_(x→0) ((sin 2x)/x) ] =−(1/(2π)) [ 0−2 ]=(1/π)

$$\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\pi−{x}\right)−\mathrm{sin}\:\mathrm{2}\left(\pi−{x}\right)}{\left({x}+\pi\right)\left({x}−\pi\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}−\mathrm{sin}\:\mathrm{2}{x}}{−{x}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:\left[\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}}\:−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{{x}}\:\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:\left[\:\mathrm{0}−\mathrm{2}\:\right]=\frac{\mathrm{1}}{\pi} \\ $$

Commented by MikeH last updated on 16/Feb/23

thanks

$$\mathrm{thanks}\: \\ $$

Answered by Frix last updated on 15/Feb/23

Simply use l′Ho^� pital ⇒ answer is (1/π)

$$\mathrm{Simply}\:\mathrm{use}\:\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\pi} \\ $$

Answered by CElcedricjunior last updated on 15/Feb/23

lim_(x→𝛑) ((cosx+sin(2x)+1)/(x^2 −𝛑^2 ))=(0/0)=FI ■Moivre lim_(x→𝛑) ((−sinx+2cos(2x))/(2x))=(2/(2𝛑))=(1/𝛑) ★cedric junior B)(1/𝛑)

$$\underset{{x}\rightarrow\boldsymbol{\pi}} {\mathrm{lim}}\frac{\boldsymbol{{cosx}}+\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)+\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{\pi}^{\mathrm{2}} }=\frac{\mathrm{0}}{\mathrm{0}}=\boldsymbol{{F}\mathrm{I}}\:\blacksquare\boldsymbol{{M}}{oi}\boldsymbol{{vre}} \\ $$$$\underset{{x}\rightarrow\boldsymbol{\pi}} {\mathrm{lim}}\frac{−\boldsymbol{{sinx}}+\mathrm{2}\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)}{\mathrm{2}\boldsymbol{{x}}}=\frac{\mathrm{2}}{\mathrm{2}\boldsymbol{\pi}}=\frac{\mathrm{1}}{\boldsymbol{\pi}}\:\bigstar{cedric}\:{junior} \\ $$$$\left.\boldsymbol{{B}}\right)\frac{\mathrm{1}}{\boldsymbol{\pi}} \\ $$

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