lim-x-2-f-x-2-6-x-2-4-if-lim-x-4-4-f-x-24-x-4-

Question Number 126824 by MathSh last updated on 24/Dec/20

$$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{f}\left({x}+\mathrm{2}\right)−\mathrm{6}}{{x}−\mathrm{2}}=\mathrm{4}\:\:{if},\:\:\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\mathrm{4}\centerdot{f}\left({x}\right)−\mathrm{24}}{{x}−\mathrm{4}}=? \\ $$

Answered by mahdipoor last updated on 24/Dec/20

lim_(x→2) ((f(x+2)−6)/(x−2))=((f(4)−6)/0)=4⇒f(4)−6=0 ⇒^(L′hopital ) lim_(x→2) ((f(x+2)−6)/(x−2))=((f ′(x+2))/1)=f ′(4)=4 lim_(x→4) ((4∙f(x)−24)/(x−4))=(0/0)⇒^(L′hopital) ((4×f ′(x))/1)=4×f ′(4)=16

$$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{f}\left({x}+\mathrm{2}\right)−\mathrm{6}}{{x}−\mathrm{2}}=\frac{{f}\left(\mathrm{4}\right)−\mathrm{6}}{\mathrm{0}}=\mathrm{4}\Rightarrow{f}\left(\mathrm{4}\right)−\mathrm{6}=\mathrm{0} \\ $$$$\overset{{L}'{hopital}\:} {\Rightarrow}\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{f}\left({x}+\mathrm{2}\right)−\mathrm{6}}{{x}−\mathrm{2}}=\frac{{f}\:'\left({x}+\mathrm{2}\right)}{\mathrm{1}}={f}\:'\left(\mathrm{4}\right)=\mathrm{4} \\ $$$$\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\mathrm{4}\centerdot{f}\left({x}\right)−\mathrm{24}}{{x}−\mathrm{4}}=\frac{\mathrm{0}}{\mathrm{0}}\overset{{L}'{hopital}} {\Rightarrow}\frac{\mathrm{4}×{f}\:'\left({x}\right)}{\mathrm{1}}=\mathrm{4}×{f}\:'\left(\mathrm{4}\right)=\mathrm{16} \\ $$

Commented by MathSh last updated on 24/Dec/20