Question Number 168555 by mathlove last updated on 13/Apr/22
Answered by LEKOUMA last updated on 13/Apr/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{4}} }{\mathrm{3}}+{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}^{\mathrm{4}} }{\mathrm{3}{x}^{\mathrm{4}} }=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by qaz last updated on 13/Apr/22
![ln(1−x)ln(1+x)+x^2 =−(x+(1/2)x^2 +(1/3)x^3 +...)(x−(1/2)x^2 +(1/3)x^3 +...)+x^2 =−[(x+(1/3)x^3 )^2 −(1/4)x^4 +...]+x^2 =−(x^2 +(2/3)x^4 −(1/4)x^4 +...)+x^2 =−(5/(12))x^4 +o(x^4 ) ⇒lim_(x→0) ((ln(1−x)ln(1+x)+x^2 )/x^4 )=lim_(x−0) ((−(5/(12))x^4 +o(x^4 ))/x^4 )=−(5/(12))](https://www.tinkutara.com/question/Q168574.png)
$$\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)+\mathrm{x}^{\mathrm{2}} =−\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} +…\right)\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} +…\right)+\mathrm{x}^{\mathrm{2}} \\ $$$$=−\left[\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{4}} +…\right]+\mathrm{x}^{\mathrm{2}} =−\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{4}} +…\right)+\mathrm{x}^{\mathrm{2}} =−\frac{\mathrm{5}}{\mathrm{12}}\mathrm{x}^{\mathrm{4}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)+\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} }=\underset{\mathrm{x}−\mathrm{0}} {\mathrm{lim}}\frac{−\frac{\mathrm{5}}{\mathrm{12}}\mathrm{x}^{\mathrm{4}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{x}^{\mathrm{4}} }=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$
Commented by mathlove last updated on 17/Apr/22
$${thanks} \\ $$