Question Number 144204 by bemath last updated on 23/Jun/21
$$\mathrm{Find}\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{f}\left(\mathrm{2h}+\mathrm{2}+\mathrm{h}^{\mathrm{2}} \right)−\mathrm{f}\left(\mathrm{2}\right)}{\mathrm{f}\left(\mathrm{h}−\mathrm{h}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{f}\left(\mathrm{1}\right)}=? \\ $$$$\mathrm{if}\:\mathrm{given}\:\mathrm{that}\:\begin{cases}{\mathrm{f}\:'\left(\mathrm{2}\right)=\mathrm{6}}\\{\mathrm{f}\:'\left(\mathrm{1}\right)=\mathrm{4}}\end{cases} \\ $$
Answered by bramlexs22 last updated on 23/Jun/21
$$\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}+\mathrm{2h}\right)\mathrm{f}\:'\left(\mathrm{2h}+\mathrm{2}+\mathrm{h}^{\mathrm{2}} \right)}{\left(\mathrm{1}−\mathrm{2h}\right)\mathrm{f}\:'\left(\mathrm{h}−\mathrm{h}^{\mathrm{2}} +\mathrm{1}\right)}\: \\ $$$$=\:\frac{\mathrm{2f}\:'\left(\mathrm{2}\right)}{\mathrm{f}\:'\left(\mathrm{1}\right)}\:=\:\frac{\mathrm{12}}{\mathrm{4}}=\mathrm{3} \\ $$