Let $f_n=\left(1+\frac{1}{n}\right)^n\left((2n-1)!F_n\right)^{\frac{1}{n}}$. Find $\lim \limits_{n \to \infty}(f_{n+1} - f_n)$ where $F_n$ denotes the $n$th Fibonacci number (given by $F_0 = 0$, $F_1 = 1$, and by $F_{n+1} = F_n + F_{n-1}$ for all $n \geq 1$
To find the limit \(\lim_{n \to \infty}(f_{n+1} - f_n)\), we start by analyzing the given expression for \(f_n\):

\[ f_n = \left(1 + \frac{1}{n}\right)^n \left((2n-1)! F_n\right)^{\frac{1}{n}} \]

1. **Examine the first part of \(f_n\):**
   \[ \left(1 + \frac{1}{n}\right)^n \]
   As \(n \to \infty\), we know from the definition of the exponential function that:
   \[ \left(1 + \frac{1}{n}\right)^n \to e \]

2. **Examine the second part of \(f_n\):**
   \[ \left((2n-1)! F_n\right)^{\frac{1}{n}} \]
   We need to understand the behavior of \((2n-1)!\) and \(F_n\) as \(n \to \infty\).

3. **Stirling's approximation for factorials:**
   Using Stirling's approximation for large \(n\):
   \[ (2n-1)! \approx \sqrt{2\pi(2n-1)} \left(\frac{2n-1}{e}\right)^{2n-1} \]
   Therefore:
   \[ ((2n-1)!)^{\frac{1}{n}} \approx \left(\sqrt{2\pi(2n-1)} \left(\frac{2n-1}{e}\right)^{2n-1}\right)^{\frac{1}{n}} \]
   Simplifying this:
   \[ ((2n-1)!)^{\frac{1}{n}} \approx \left(2\pi(2n-1)\right)^{\frac{1}{2n}} \left(\frac{2n-1}{e}\right)^{2 - \frac{1}{n}} \]
   As \(n \to \infty\), \(\left(2\pi(2n-1)\right)^{\frac{1}{2n}} \to 1\) and \(\left(\frac{2n-1}{e}\right)^{2 - \frac{1}{n}} \approx \left(\frac{2n-1}{e}\right)^2\).

4. **Behavior of Fibonacci numbers:**
   The \(n\)-th Fibonacci number \(F_n\) grows asymptotically as:
   \[ F_n \approx \frac{\phi^n}{\sqrt{5}} \]
   where \(\phi = \frac{1 + \sqrt{5}}{2}\) is the golden ratio.

5. **Combining the parts:**
   \[ \left((2n-1)! F_n\right)^{\frac{1}{n}} \approx \left(\left(\frac{2n-1}{e}\right)^2 \frac{\phi^n}{\sqrt{5}}\right)^{\frac{1}{n}} \]
   Simplifying:
   \[ \left((2n-1)! F_n\right)^{\frac{1}{n}} \approx \left(\frac{(2n-1)^2}{e^2} \cdot \frac{\phi^n}{\sqrt{5}}\right)^{\frac{1}{n}} \]
   \[ \approx \left(\frac{(2n-1)^2}{e^2}\right)^{\frac{1}{n}} \left(\frac{\phi^n}{\sqrt{5}}\right)^{\frac{1}{n}} \]
   \[ \approx \left(\frac{2n-1}{e}\right)^{\frac{2}{n}} \phi \left(\frac{1}{\sqrt{5}}\right)^{\frac{1}{n}} \]
   As \(n \to \infty\), \(\left(\frac{2n-1}{e}\right)^{\frac{2}{n}} \to 1\) and \(\left(\frac{1}{\sqrt{5}}\right)^{\frac{1}{n}} \to 1\), so:
   \[ \left((2n-1)! F_n\right)^{\frac{1}{n}} \approx \phi \]

6. **Combining all parts:**
   \[ f_n \approx e \cdot \phi \]

7. **Finding the limit:**
   \[ \lim_{n \to \infty} (f_{n+1} - f_n) = \lim_{n \to \infty} (e \cdot \phi - e \cdot \phi) = 0 \]

\(\blacksquare\)

The final answer is \( \boxed{ 0 } \).