The expression given is:

$\lim_{{x \to \infty}} \left( \frac{x + 6}{x} \right)^x$

Let's solve it step-by-step:

Step 1: Simplify the Expression Inside the Limit

$\frac{x + 6}{x} = \frac{x}{x} + \frac{6}{x} = 1 + \frac{6}{x}$

So, the limit now becomes: $\lim_{{x \to \infty}} \left( 1 + \frac{6}{x} \right)^x$

Step 2: Use the Exponential Limit Property

Recall that for limits of the form $\lim_{{x \to \infty}} \left( 1 + \frac{a}{x} \right)^x$, the limit is $e^a$. Here, $a = 6$.

Thus: $\lim_{{x \to \infty}} \left( 1 + \frac{6}{x} \right)^x = e^6$

Final Answer

$\lim_{{x \to \infty}} \left( \frac{x + 6}{x} \right)^x = e^6$

If you'd like further details or have any questions, feel free to ask!

Related Questions:

1. What is the exponential limit rule for $\lim_{{x \to \infty}} \left( 1 + \frac{a}{x} \right)^x$ and why does it work?
2. How does this problem change if the numerator had a different constant, e.g., $x + 12$?
3. What are some other forms of limits that approach $e$?
4. How would you solve the limit if the expression were $\lim_{{x \to \infty}} \left( 1 - \frac{6}{x} \right)^x$?
5. What techniques can be used to evaluate limits that result in indeterminate forms like $1^\infty$?

Tip:

When evaluating limits of the form $\left( 1 + \frac{a}{x} \right)^x$ as $x \to \infty$, always remember that it converges to $e^a$.