Question Number 84333 by M±th+et£s last updated on 11/Mar/20
$$\left.\mathrm{1}\right){find}\:{without}\:{l}'{hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{2}\sqrt{{x}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}}{{x}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{the}\:{general}\:{solution}\:{for}\:{tbe}\:{differential}\:{equation} \\ $$$$\left(\mathrm{1}+{y}^{\mathrm{2}} \right)+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\frac{{dy}}{{dx}}\right)=\mathrm{0}\:{is}\:{y}=\frac{{k}−{x}}{\mathrm{1}+{kx}},{k}\:{is}\:{a}\:{constant} \\ $$$${then}\:{find}\:{the}\:{special}\:{solution}\:{if}\:{y}=\frac{\mathrm{2}}{\mathrm{3}\:}\:{when}\:{x}=\mathrm{1} \\ $$
Answered by mind is power last updated on 11/Mar/20
$$\left(\mathrm{1}+{x}\right)^{{a}} =\mathrm{1}+{ax}+{o}\left({x}\right)\Rightarrow \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{1}+{x}}−\sqrt[{\mathrm{3}}]{\mathrm{1}+{x}}−\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}}{{x}}=\frac{\mathrm{2}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}+{o}\left({x}\right)\right)−\left(\mathrm{1}+\frac{{x}}{\mathrm{3}}+{o}\left({x}\right)\right)−\left(\mathrm{1}+\frac{{x}}{\mathrm{4}}+{o}\left({x}\right)\right)}{{x}} \\ $$$$=\frac{\frac{\mathrm{5}{x}}{\mathrm{12}}+{o}\left({x}\right)}{{x}}=\frac{\mathrm{5}}{\mathrm{12}}+{o}\left(\mathrm{1}\right)\rightarrow\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$ \\ $$
Answered by mind is power last updated on 11/Mar/20
$$\Rightarrow\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }=\frac{−{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{arctan}\left({y}\right)=−{arctan}\left({x}\right)+{c}\Rightarrow \\ $$$${y}={tg}\left({c}−{arctan}\left({x}\right)\right)=\frac{{tg}\left({c}\right)−{tg}\left({arctan}\left({x}\right)\right)}{\mathrm{1}+{tg}\left({c}\right){tg}\left({arctan}\left({x}\right)\right)} \\ $$$$\Rightarrow{y}=\frac{{tg}\left({c}\right)−{x}}{\mathrm{1}+{tg}\left({c}\right).{x}},{let}\:{tg}\left({c}\right)={k}\Rightarrow \\ $$$${y}=\frac{{k}−{x}}{\mathrm{1}+{kx}} \\ $$$${x}=\mathrm{1}\Rightarrow{y}=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+{k}\right)={k}−\mathrm{1}\Rightarrow{k}=\mathrm{5} \\ $$$${y}\left({x}\right)=\frac{\mathrm{5}−{x}}{\mathrm{1}+\mathrm{5}{x}} \\ $$
Commented by M±th+et£s last updated on 11/Mar/20
$${god}\:{bless}\:{you}\:{sir} \\ $$
Answered by behi83417@gmail.com last updated on 11/Mar/20
$$\mathrm{x}+\mathrm{1}=\mathrm{t}^{\mathrm{12}} ,\mathrm{x}\rightarrow\mathrm{0}\Rightarrow\mathrm{t}\rightarrow\mathrm{1} \\ $$$$\mathrm{L}=\underset{\mathrm{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2t}^{\mathrm{6}} −\mathrm{t}^{\mathrm{4}} −\mathrm{t}^{\mathrm{3}} }{\mathrm{t}^{\mathrm{12}} −\mathrm{1}}=\underset{\mathrm{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{12t}^{\mathrm{5}} −\mathrm{4t}^{\mathrm{3}} −\mathrm{3t}^{\mathrm{2}} }{\mathrm{12t}^{\mathrm{11}} }= \\ $$$$=\frac{\mathrm{12}−\mathrm{4}−\mathrm{3}}{\mathrm{12}}=\frac{\mathrm{5}}{\mathrm{12}}\:\:\:.\blacksquare \\ $$