Question Number 27701 by NECx last updated on 13/Jan/18
$${If}\:{the}\:{function}\:{f}\left({x}\right)\:{satisfies} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\frac{{f}\left({x}\right)−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\pi,\:{evaluate}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right) \\ $$
Commented by abdo imad last updated on 21/Jan/18
$$\Leftrightarrow\:{lim}_{{x}\rightarrow\mathrm{1}} \:\:\:\:\:\frac{{f}\left({x}\right)−\mathrm{2}\:−\pi\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{0}\:{so}\:{we}\:{must}\:{have} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} {f}\left({x}\right)−\mathrm{2}−\pi\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0}\:{in}\:{order}\:{to}\:{find}\:{the}\:{form}\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{1}} {f}\left({x}\right)=\mathrm{2}\:\:\:{let}\:{verify}\:{this}\:{number}\:{by}\:{hospital}\: \\ $$$${theorem}\:{lim}_{{x}\rightarrow\mathrm{1}} \frac{{f}^{'} \left({x}\right)\:−\mathrm{2}\pi{x}}{\mathrm{2}{x}}=\mathrm{0}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}} {f}^{'} \left({x}\right)=\mathrm{2}\pi\:{so} \\ $$$${we}\:{must}\:{have}\:{lim}_{{x}\rightarrow\mathrm{1}} \:{f}\left({x}\right)=\mathrm{2}\:{and}\:{lim}_{{x}\rightarrow\mathrm{1}} {f}^{'} \left({x}\right)=\mathrm{2}\pi\:. \\ $$
Answered by mrW2 last updated on 13/Jan/18
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{2} \\ $$
Commented by NECx last updated on 13/Jan/18
$${how}? \\ $$
Commented by mrW2 last updated on 14/Jan/18
$${if}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:{f}\left({x}\right)\neq\mathrm{2}, \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{finite}}{\mathrm{0}}\rightarrow\infty\neq\pi \\ $$
Commented by abdo imad last updated on 21/Jan/18
$${this}\:{condition}\:{is}\:{unsufficient}.... \\ $$