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Question Number 133738 by mathlove last updated on 23/Feb/21

Answered by Ar Brandon last updated on 24/Feb/21

$$\underset{\mathrm{x}\to 0}{\lim }\frac{\ln (3\mathrm{x}+1)}{2\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{3\mathrm{x}-\frac{{9\mathrm{x}}^2}{2}+ϵ\left(\mathrm{x}\right)}{2\mathrm{x}}=\frac{3}{2}$$

Commented by mathlove last updated on 24/Feb/21

$$whattheϵ\left(x\right)=?$$$$$$

Commented by Ar Brandon last updated on 24/Feb/21

$$\ln (1+\mathrm{x})=\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^4}{4}+\frac{{\mathrm{x}}^5}{5}+\cdot \cdot \cdot +\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}+\cdot \cdot \cdot$$$$\ln (1+3\mathrm{x})=3\mathrm{x}-\frac{{9\mathrm{x}}^2}{2}+\frac{{27\mathrm{x}}^3}{3}-\frac{{81\mathrm{x}}^4}{4}+\frac{{243\mathrm{x}}^5}{5}+\cdot \cdot \cdot +\frac{3^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}+\cdot \cdot \cdot$$

Answered by bramlexs22 last updated on 24/Feb/21

$$10.\underset{x\to 0}{\lim }\frac{\ln (3\mathrm{x}+1)}{2\mathrm{x}}=\underset{x\to 0}{\lim }\frac{\left[\frac{3}{3\mathrm{x}+1}\right]}{2}=\frac{3}{2}$$

Answered by Dwaipayan Shikari last updated on 24/Feb/21

$$\underset{x\to 0}{\lim }\frac{log(1+3x)}{2x}\underset{x\to 0}{\lim }log(1+x)=x$$$$=\frac{3x}{2x}=\frac{3}{2}$$

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