6.120 . lim _(xarrow 0)((1+3x)/(1+x))^(5)/(x)

To evaluate the limit \(\lim_{x \to 0} \left( \frac{1 + 3x}{1 + x} \right)^{\frac{5}{x}}\), we can use the property of limits involving exponentials and logarithms. Let's denote the expression inside the limit as \(L\):<br /><br />\[ L = \left( \frac{1 + 3x}{1 + x} \right)^{\frac{5}{x}} \]<br /><br />Taking the natural logarithm of both sides, we get:<br /><br />\[ \ln L = \ln \left( \left( \frac{1 + 3x}{1 + x} \right)^{\frac{5}{x}} \right) \]<br /><br />Using the property of logarithms, we can bring the exponent in front:<br /><br />\[ \ln L = \frac{5}{x} \ln \left( \frac{1 + 3x}{1 + x} \right) \]<br /><br />Now, we need to evaluate the limit of the right-hand side as \(x\) approaches 0:<br /><br />\[ \lim_{x \to 0} \frac{5}{x} \ln \left( \frac{1 + 3x}{1 + x} \right) \]<br /><br />We can simplify the argument of the logarithm:<br /><br />\[ \frac{1 + 3x}{1 + x} = \frac{(1 + x) + 2x}{1 + x} = 1 + \frac{2x}{1 + x} \]<br /><br />As \(x\) approaches 0, \(\frac{2x}{1 + x}\) approaches 0, so:<br /><br />\[ \ln \left( \frac{1 + 3x}{1 + x} \right) \approx \ln (1 + \frac{2x}{1 + x}) \approx \frac{2x}{1 + x} \]<br /><br />Thus, the expression becomes:<br /><br />\[ \lim_{x \to 0} \frac{5}{x} \cdot \frac{2x}{1 + x} = \lim_{x \to 0} \frac{10}{1 + x} = 10 \]<br /><br />Therefore, the natural logarithm of the limit \(L\) is 10:<br /><br />\[ \ln L = 10 \]<br /><br />Exponentiating both sides to solve for \(L\):<br /><br />\[ L = e^{10} \]<br /><br />So, the limit is:<br /><br />\[ \lim_{x \to 0} \left( \frac{1 + 3x}{1 + x} \right)^{\frac{5}{x}} = e^{10} \]