Question Number 190765 by mustafazaheen last updated on 10/Apr/23
Answered by qaz last updated on 11/Apr/23
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−{e}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{2}{x}\right)^{\frac{\mathrm{1}}{{x}}} }{\mathrm{sin}\:{x}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−{e}^{\mathrm{2}} +{e}^{\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)} }{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−{e}^{\mathrm{2}} +{e}^{\mathrm{2}−\mathrm{2}{x}+{o}\left({x}\right)} }{{x}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−{e}^{\mathrm{2}} +{e}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{x}+{o}\left({x}\right)\right)}{{x}}=−\mathrm{2}{e}^{\mathrm{2}} \\ $$
Commented by mustafazaheen last updated on 12/Apr/23
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$$$\mathrm{o}\left(\mathrm{x}\right)=? \\ $$