Determine the limit of the given function as x approaches 0: $\lim _{x \rightarrow 0}\left(\frac{7+4 x}{7}\right)^{\frac{-21}{4 x}}$

Step 1 ：Define the function as \(f = e^{-\frac{21}{4x} \cdot \ln(\frac{7+4x}{7})}\)

Step 2 ：Differentiate the numerator to get \(\frac{d}{dx}(-\frac{21}{4x}) = \frac{21}{4x^2}\)

Step 3 ：Differentiate the denominator to get \(\frac{d}{dx}(\ln(\frac{7+4x}{7})) = \frac{4}{7(\frac{4x}{7} + 1)}\)

Step 4 ：Apply L'Hopital's rule to find the limit as x approaches 0, which gives \(\lim_{x \to 0} \frac{\frac{21}{4x^2}}{\frac{4}{7(\frac{4x}{7} + 1)}} = \infty\)

Step 5 ：This means that as x approaches 0, the value of the function increases without bound.

Step 6 ：Final Answer: The limit of the function as x approaches 0 is \(\boxed{\infty}\)