Question Number 183927 by greougoury555 last updated on 31/Dec/22
$$\:\underset{{x}\rightarrow\mathrm{2}\:} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{x}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{12}}\:=? \\ $$
Answered by cortano1 last updated on 31/Dec/22
![lim_(x→2) ((tan (2πx)+cos ((π/2)x)+tan ((π/8)x))/((x−2)(x+6))) = lim_(x→2) ((tan (2πx)+cos ((π/2)x)+tan ((π/8)x))/(8(x−2))) [ x−2=m ] = lim_(m→0) ((tan (2π(m+2))+cos ((π/2)(m+2))+tan ((π/8)(m+2)))/(8m)) =lim_(m→0) ((tan (2πm)−cos ((π/2)m)+tan ((π/8)m+(π/4)))/(8m)) =(π/4)+lim_(m→0) ((1−cos ((π/2)m))/(8m)) +lim_(m→0) ((tan ((π/8)m+(π/4))−tan (π/4))/(8m)) =(π/4)+lim_(m→0) ((sin^2 ((π/2)m))/(16m)) +lim_(m→0) ((tan ((π/8)m)[ 1+tan ((π/8)m+(π/4))])/(8m)) =(π/4)+0+((2.π)/(64)) = ((9π)/(32))](https://www.tinkutara.com/question/Q183928.png)
$$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{x}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{x}\right)}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{6}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{x}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{x}\right)}{\mathrm{8}\left({x}−\mathrm{2}\right)} \\ $$$$\left[\:{x}−\mathrm{2}={m}\:\right] \\ $$$$=\:\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi\left({m}+\mathrm{2}\right)\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}\left({m}+\mathrm{2}\right)\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}\left({m}+\mathrm{2}\right)\right)}{\mathrm{8}{m}} \\ $$$$=\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{m}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{m}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{m}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{8}{m}} \\ $$$$=\frac{\pi}{\mathrm{4}}+\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{m}\right)}{\mathrm{8}{m}}\:+\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{m}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{tan}\:\frac{\pi}{\mathrm{4}}}{\mathrm{8}{m}} \\ $$$$=\frac{\pi}{\mathrm{4}}+\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{m}\right)}{\mathrm{16}{m}}\:+\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{m}\right)\left[\:\mathrm{1}+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{m}+\frac{\pi}{\mathrm{4}}\right)\right]}{\mathrm{8}{m}} \\ $$$$=\frac{\pi}{\mathrm{4}}+\mathrm{0}+\frac{\mathrm{2}.\pi}{\mathrm{64}}\:=\:\frac{\mathrm{9}\pi}{\mathrm{32}}\: \\ $$
Answered by qaz last updated on 31/Dec/22
$$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{x}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{12}} \\ $$$$=\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\mathrm{2}\pi\mathrm{sec}\:^{\mathrm{2}} \left(\mathrm{2}\pi{x}\right)−\frac{\pi}{\mathrm{2}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\frac{\pi}{\mathrm{8}}\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}{x}\right)}{\mathrm{2}{x}+\mathrm{4}} \\ $$$$=\frac{\mathrm{2}\pi−\mathrm{0}+\frac{\pi}{\mathrm{8}}\centerdot\mathrm{2}}{\mathrm{8}}=\frac{\mathrm{9}}{\mathrm{32}}\pi \\ $$