The $n^{th}$ derivative of a function $f(x)$ (if it exists) is denoted by $f^{(n)}(x) $. Let $f(x)=\frac{e^x}{x}$. 
Suppose $f$ is differentiable infinitely many times in $(0,\infty) $. Then find $\lim_{n \to \infty}\frac{f^{(2n)}1}{(2n)!}$
1. Given the function \( f(x) = \frac{e^x}{x} \), we need to find the limit \(\lim_{n \to \infty} \frac{f^{(2n)}(1)}{(2n)!}\).

2. First, we express \( f(x) \) using the Maclaurin series expansion for \( e^x \):
   \[
   e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}
   \]
   Therefore,
   \[
   f(x) = \frac{e^x}{x} = \frac{1}{x} \sum_{k=0}^{\infty} \frac{x^k}{k!} = \sum_{k=0}^{\infty} \frac{x^{k-1}}{k!}
   \]

3. To find the \( n \)-th derivative of \( f(x) \), we note that:
   \[
   f^{(n)}(x) = \sum_{k=0}^{\infty} \frac{(k-1)(k-2)\cdots(k-n)}{k!} x^{k-n-1}
   \]
   Evaluating this at \( x = 1 \):
   \[
   f^{(n)}(1) = \sum_{k=0}^{\infty} \frac{(k-1)(k-2)\cdots(k-n)}{k!}
   \]

4. For large \( n \), the dominant term in \( f^{(2n)}(1) \) will be the one corresponding to \( k = 2n \):
   \[
   f^{(2n)}(1) \approx \frac{(2n-1)(2n-2)\cdots(1)}{(2n)!} = \frac{(2n)!}{(2n)!} = 1
   \]

5. Therefore, we have:
   \[
   \frac{f^{(2n)}(1)}{(2n)!} \approx \frac{1}{(2n)!}
   \]

6. Taking the limit as \( n \to \infty \):
   \[
   \lim_{n \to \infty} \frac{f^{(2n)}(1)}{(2n)!} = \lim_{n \to \infty} \left( 1 + \frac{1}{(2n)!} \left[ \frac{1}{2n+1} + \frac{1}{2n+2} + \frac{1}{(2n+3) \cdot 2} + \cdots \right] \right) = 1
   \]

The final answer is \( \boxed{ 1 } \)