Let's evaluate the limit of the following expression as $x \to 3$:

$\lim_{x \to 3} \left( (4 - x)^{\frac{1}{x}} - 3 \right)$

Step 1: Substituting $x = 3$

First, let's substitute $x = 3$ directly into the expression to see if we get any indeterminate form.

$(4 - 3)^{\frac{1}{3}} - 3 = 1^{\frac{1}{3}} - 3 = 1 - 3 = -2$

Step 2: Conclusion

Since substituting $x = 3$ does not lead to an indeterminate form and directly evaluates to $-2$, we conclude that:

$\lim_{x \to 3} \left( (4 - x)^{\frac{1}{x}} - 3 \right) = -2$

Let me know if you'd like a further explanation or if you have more questions about this problem.

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Here are five related questions to explore further:

1. How do limits behave when exponents approach 1?
2. How would the result change if the expression was $(4 - x)^{\frac{1}{x}} + 3$?
3. What happens when the base of an exponential expression approaches 0?
4. How can you evaluate limits involving powers like $(4 - x)^{\frac{1}{x}}$ using L'Hopital's Rule?
5. How do you handle limits involving roots or fractional exponents?

Tip: Always check if direct substitution leads to an indeterminate form like $0/0$ or $\infty/\infty$ before considering more advanced limit evaluation techniques.