1. **State the problem:** Find the limit $$\lim_{x \to 2} \left(x^3 - 4x + 1\right)^{\frac{1}{x^2 - 4}}.$$\n\n2. **Identify the form:** Substitute $x=2$ directly: $$2^3 - 4 \cdot 2 + 1 = 8 - 8 + 1 = 1,$$ and $$2^2 - 4 = 4 - 4 = 0,$$ so the expression is of the form $1^{\infty}$, an indeterminate form. \n3. **Use logarithms to handle the limit:** Let $$y = \left(x^3 - 4x + 1\right)^{\frac{1}{x^2 - 4}}.$$ Taking natural logarithm, $$\ln y = \frac{\ln\left(x^3 - 4x + 1\right)}{x^2 - 4}.$$\nWe want $$\lim_{x \to 2} \ln y = \lim_{x \to 2} \frac{\ln\left(x^3 - 4x + 1\right)}{x^2 - 4}.$$\n\n4. **Apply substitution:** Let $$f(x) = \ln\left(x^3 - 4x + 1\right), \quad g(x) = x^2 - 4.$$\nAs $x \to 2$, both numerator and denominator approach 0, so we can apply L'Hôpital's Rule. \n5. **Differentiate numerator and denominator:** $$f'(x) = \frac{3x^2 - 4}{x^3 - 4x + 1}, \quad g'(x) = 2x.$$\n\n6. **Evaluate the limit using L'Hôpital's Rule:** $$\lim_{x \to 2} \frac{f'(x)}{g'(x)} = \lim_{x \to 2} \frac{\frac{3x^2 - 4}{x^3 - 4x + 1}}{2x} = \lim_{x \to 2} \frac{3x^2 - 4}{2x(x^3 - 4x + 1)}.$$\nSubstitute $x=2$: $$\frac{3(2)^2 - 4}{2 \cdot 2 \cdot (8 - 8 + 1)} = \frac{3 \cdot 4 - 4}{4 \cdot 1} = \frac{12 - 4}{4} = \frac{8}{4} = 2.$$\n\n7. **Find the original limit:** Since $$\lim_{x \to 2} \ln y = 2,$$ then $$\lim_{x \to 2} y = e^2.$$\n\n**Final answer:** $$\boxed{e^2}.$$