lim _(xarrow 0)(ln(1+x))/(3x)

To evaluate the limit \(\lim_{x \rightarrow 0} \frac{\ln(1+x)}{3x}\), we can use the fact that \(\ln(1+x)\) can be expanded into a Taylor series around \(x = 0\). The Taylor series expansion for \(\ln(1+x)\) is:<br /><br />\[<br />\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots<br />\]<br /><br />We only need the first-order term because higher-order terms will become negligible as \(x\) approaches 0. Therefore, we can approximate \(\ln(1+x)\) by \(x\) near \(x = 0\):<br /><br />\[<br />\ln(1+x) \approx x<br />\]<br /><br />Substituting this approximation into the limit, we get:<br /><br />\[<br />\lim_{x \rightarrow 0} \frac{\ln(1+x)}{3x} \approx \lim_{x \rightarrow 0} \frac{x}{3x}<br />\]<br /><br />The \(x\) terms cancel out:<br /><br />\[<br />\lim_{x \rightarrow 0} \frac{x}{3x} = \lim_{x \rightarrow 0} \frac{1}{3} = \frac{1}{3}<br />Therefore, the limit is:<br /><br />\[<br />\lim_{x \rightarrow 0} \frac{\ln(1+x)}{3x} = \frac{1}{3}<br />\]