To solve $\lim_{x \to \infty} \left(x + \frac{1}{x}\right)^x$, we can use the property of limits involving exponential functions.

First, rewrite the expression:

$\left(x + \frac{1}{x}\right)^x = \left(x\left(1 + \frac{1}{x^2}\right)\right)^x = x^x \left(1 + \frac{1}{x^2}\right)^x$

Now, consider the individual limits:

1. $x^x$ goes to $\infty$ as $x \to \infty$.
2. For $\left(1 + \frac{1}{x^2}\right)^x$, take the natural log:

$\ln\left[\left(1 + \frac{1}{x^2}\right)^x\right] = x \cdot \ln\left(1 + \frac{1}{x^2}\right)$

Using the expansion $\ln(1 + u) \approx u$ for small $u$, we have:

$x \cdot \ln\left(1 + \frac{1}{x^2}\right) \approx x \cdot \frac{1}{x^2} = \frac{1}{x}$

As $x \to \infty$, $\frac{1}{x} \to 0$.

Hence:

$\lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^x = e^0 = 1$

Putting it all together, the dominant term is $x^x$, which results in:

$\lim_{x \to \infty} \left(x + \frac{1}{x}\right)^x \to \infty$