The sequence $\{c_{n}\}$ is determined by the following equation. \[c_{n}=(n+1)\int_{0}^{1}x^{n}\cos \pi x\ dx\ (n=1,\ 2,\ \cdots).\] Let $\lambda$ be the limit value $\lim_{n\to\infty}c_{n}.$  Find $\lim_{n\to\infty}\frac{c_{n+1}-\lambda}{c_{n}-\lambda}.$
1. We start with the given sequence:
   \[
   c_{n} = (n+1) \int_{0}^{1} x^{n} \cos(\pi x) \, dx
   \]
   
2. To evaluate the integral, we use integration by parts. Let \( u = x^n \) and \( dv = \cos(\pi x) \, dx \). Then, \( du = n x^{n-1} \, dx \) and \( v = \frac{\sin(\pi x)}{\pi} \). Applying integration by parts:
   \[
   \int_{0}^{1} x^n \cos(\pi x) \, dx = \left. \frac{x^n \sin(\pi x)}{\pi} \right|_{0}^{1} - \int_{0}^{1} \frac{n x^{n-1} \sin(\pi x)}{\pi} \, dx
   \]
   Evaluating the boundary term:
   \[
   \left. \frac{x^n \sin(\pi x)}{\pi} \right|_{0}^{1} = \frac{\sin(\pi)}{\pi} - \frac{0}{\pi} = 0
   \]
   Thus, we have:
   \[
   \int_{0}^{1} x^n \cos(\pi x) \, dx = -\frac{n}{\pi} \int_{0}^{1} x^{n-1} \sin(\pi x) \, dx
   \]

3. We now need to evaluate the integral involving \(\sin(\pi x)\). Using integration by parts again, let \( u = x^{n-1} \) and \( dv = \sin(\pi x) \, dx \). Then, \( du = (n-1) x^{n-2} \, dx \) and \( v = -\frac{\cos(\pi x)}{\pi} \). Applying integration by parts:
   \[
   \int_{0}^{1} x^{n-1} \sin(\pi x) \, dx = \left. -\frac{x^{n-1} \cos(\pi x)}{\pi} \right|_{0}^{1} + \int_{0}^{1} \frac{(n-1) x^{n-2} \cos(\pi x)}{\pi} \, dx
   \]
   Evaluating the boundary term:
   \[
   \left. -\frac{x^{n-1} \cos(\pi x)}{\pi} \right|_{0}^{1} = -\frac{\cos(\pi)}{\pi} + \frac{1}{\pi} = 0
   \]
   Thus, we have:
   \[
   \int_{0}^{1} x^{n-1} \sin(\pi x) \, dx = \frac{n-1}{\pi} \int_{0}^{1} x^{n-2} \cos(\pi x) \, dx
   \]

4. Substituting back, we get:
   \[
   \int_{0}^{1} x^n \cos(\pi x) \, dx = -\frac{n}{\pi} \cdot \frac{n-1}{\pi} \int_{0}^{1} x^{n-2} \cos(\pi x) \, dx
   \]
   This recursive relation suggests that:
   \[
   \int_{0}^{1} x^n \cos(\pi x) \, dx = (-1)^k \frac{n(n-1)\cdots(n-k+1)}{\pi^k} \int_{0}^{1} x^{n-k} \cos(\pi x) \, dx
   \]
   As \( n \to \infty \), the integral term tends to zero, and we can approximate:
   \[
   c_n \approx -1
   \]

5. Given that \( \lambda = -1 \), we rewrite the sequence:
   \[
   c_n + 1 = -\frac{\pi^2}{(n+2)(n+3)} c_{n+2}
   \]

6. We now need to find:
   \[
   \lim_{n \to \infty} \frac{c_{n+1} - \lambda}{c_n - \lambda} = \lim_{n \to \infty} \frac{c_{n+1} + 1}{c_n + 1}
   \]
   Using the recursive relation:
   \[
   \frac{c_{n+1} + 1}{c_n + 1} = \frac{-\frac{\pi^2}{(n+3)(n+4)} c_{n+3}}{-\frac{\pi^2}{(n+2)(n+3)} c_{n+2}} = \frac{(n+2)(n+3)}{(n+3)(n+4)} \cdot \frac{c_{n+3}}{c_{n+2}}
   \]
   As \( n \to \infty \), this ratio approaches 1:
   \[
   \lim_{n \to \infty} \frac{c_{n+1} + 1}{c_n + 1} = 1
   \]

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