Question Number 221348 by RoseAli last updated on 31/May/25
$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}−\mathrm{2}^{{x}} }{{x}−\mathrm{2}} \\ $$
Answered by mr W last updated on 31/May/25
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{2}^{{x}} \right)}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\mathrm{1}−{e}^{{x}\mathrm{ln}\:\mathrm{2}} \right)}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{1}−{x}\mathrm{ln}\:\mathrm{2}−\frac{{x}^{\mathrm{2}} \mathrm{ln}^{\mathrm{2}} \:\mathrm{2}}{\mathrm{2}!}−…\right)}{{x}} \\ $$$$=−\mathrm{4}\:\mathrm{ln}\:\mathrm{2} \\ $$
Answered by mnjuly1970 last updated on 02/Jun/25
$$\:\:\:\overset{{hopital}} {=}{lim}_{{x}\rightarrow\mathrm{2}} \frac{−\mathrm{2}^{{x}} {ln}\left(\mathrm{2}\right)}{\mathrm{1}}=−\mathrm{4}{ln}\left(\mathrm{2}\right) \\ $$