Guess the value of the limit by evaluating the function at values close to where familiar.) lim _(xarrow 0)(1+x)^1/x square

To evaluate the limit \(\lim_{x \to 0} (1+x)^{1/x}\), we can use the fact that this expression is a well-known limit in calculus. The limit of \((1+x)^{1/x}\) as \(x\) approaches 0 is the definition of the number \(e\), the base of the natural logarithm.<br /><br />To understand why this is true, consider the following steps:<br /><br />1. **Rewrite the Expression Using Exponential Functions:**<br /> \[<br /> (1+x)^{1/x} = \exp\left(\frac{\ln(1+x)}{x}\right)<br /> \]<br /><br />2. **Analyze the Argument of the Exponential Function:**<br /> We need to find the limit of \(\frac{\ln(1+x)}{x}\) as \(x\) approaches 0.<br /><br />3. **Use the Taylor Series Expansion:**<br /> For small values of \(x\), \(\ln(1+x)\) can be approximated by its Taylor series:<br /> \[<br /> \ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots<br /> \]<br /> Therefore,<br /> \[<br /> \frac{\ln(1+x)}{x} \approx \frac{x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \cdots<br /> \]<br /><br />4. **Take the Limit:**<br /> As \(x\) approaches 0, the terms involving \(x\) in the expansion go to 0:<br /> \[<br /> \lim_{x \to 0} \left(1 - \frac{x}{2} + \frac{x^2}{3} - \cdots\right) = 1<br /> \]<br /><br />5. **Combine the Results:**<br /> Since \(\frac{\ln(1+x)}{x}\) approaches 1 as \(x\) approaches 0, we have:<br /> \[<br /> \lim_{x \to 0} \exp\left(\frac{\ln(1+x)}{x}\right) = \exp(1) = e<br /> \]<br /><br />Therefore, the value of the limit is:<br />\[<br />\boxed{e}<br />\]