Question Number 153346 by mathlove last updated on 06/Sep/21
Answered by liberty last updated on 06/Sep/21
$$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)−\mathrm{4}}{{x}−\mathrm{2}}=\mathrm{15}\:\rightarrow\begin{cases}{{f}\left(\mathrm{2}\right)=\mathrm{4}}\\{{f}\:'\left(\mathrm{2}\right)=\mathrm{15}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}{x}^{\mathrm{2}} −{f}\left({x}\right)}{{x}−\mathrm{2}}\:=\infty \\ $$
Answered by Mathspace last updated on 06/Sep/21
$${by}\:{lhospital}\:{we}\:{get}\:{lim}_{{x}\rightarrow\mathrm{2}} {f}^{'} \left({x}\right)=\mathrm{15} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{2}} \frac{\mathrm{2}{x}^{\mathrm{2}} −{f}\left({x}\right)}{{x}−\mathrm{2}}={lim}_{{x}\rightarrow\mathrm{2}} \mathrm{4}{x}−{f}^{'} \left({x}\right) \\ $$$$=\mathrm{8}−{f}^{'} \left(\mathrm{2}\right)=\mathrm{8}−\mathrm{15}\:=−\mathrm{7} \\ $$
Commented by liberty last updated on 07/Sep/21
$${wrong}. \\ $$$$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}^{\mathrm{2}} −{f}\left({x}\right)}{{x}−\mathrm{2}}\:\neq\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\:{L}'{Hopital}\:{not}\:{available} \\ $$