# Question-133738

Question Number 133738 by mathlove last updated on 23/Feb/21
Answered by Ar Brandon last updated on 24/Feb/21
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{3x}+\mathrm{1}\right)}{\mathrm{2x}}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3x}−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}+\epsilon\left(\mathrm{x}\right)}{\mathrm{2x}}=\frac{\mathrm{3}}{\mathrm{2}} \\$$
Commented by mathlove last updated on 24/Feb/21
$${what}\:{the}\:\epsilon\left({x}\right)=? \\$$$$\\$$
Commented by Ar Brandon last updated on 24/Feb/21
$$\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)=\mathrm{x}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}}+\centerdot\centerdot\centerdot+\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}+\centerdot\centerdot\centerdot \\$$$$\mathrm{ln}\left(\mathrm{1}+\mathrm{3x}\right)=\mathrm{3x}−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{27x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{81x}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{243x}^{\mathrm{5}} }{\mathrm{5}}+\centerdot\centerdot\centerdot+\frac{\mathrm{3}^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} }{\mathrm{n}}+\centerdot\centerdot\centerdot \\$$
Answered by bramlexs22 last updated on 24/Feb/21
$$\mathrm{10}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{3x}+\mathrm{1}\right)}{\mathrm{2x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left[\frac{\mathrm{3}}{\mathrm{3x}+\mathrm{1}}\right]}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\$$
Answered by Dwaipayan Shikari last updated on 24/Feb/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{log}\left(\mathrm{1}+\mathrm{3}{x}\right)}{\mathrm{2}{x}}\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{log}\left(\mathrm{1}+{x}\right)={x} \\$$$$=\frac{\mathrm{3}{x}}{\mathrm{2}{x}}=\frac{\mathrm{3}}{\mathrm{2}} \\$$