a) Compute
$$\lim_{n\to\infty}n\int^1_0\left(\frac{1-x}{1+x}\right)^ndx.$$
b) Let $k\ge1$ be an integer. Compute
$$\lim_{n\to\infty}n^{k+1}\int^1_0\left(\frac{1-x}{1+x}\right)^nx^kdx.$$
### Part (a)
1. **Substitution**: Let \( t = \frac{1-x}{1+x} \). Then, we have:
   \[
   x = \frac{1-t}{1+t} \quad \text{and} \quad dx = -\frac{2}{(1+t)^2} dt.
   \]
2. **Transform the integral**: Substitute \( x \) and \( dx \) in the integral:
   \[
   n \int_0^1 \left( \frac{1-x}{1+x} \right)^n dx = n \int_0^1 t^n \left( -\frac{2}{(1+t)^2} \right) dt = 2n \int_0^1 \frac{t^n}{(1+t)^2} dt.
   \]
3. **Apply the Fact**: Using the given fact with \( f(t) = \frac{2}{(1+t)^2} \), we get:
   \[
   \lim_{n \to \infty} n \int_0^1 t^n f(t) \, dt = f(1) = \frac{2}{(1+1)^2} = \frac{1}{2}.
   \]

### Part (b)
1. **Substitution**: Again, let \( t = \frac{1-x}{1+x} \). Then, we have:
   \[
   x = \frac{1-t}{1+t} \quad \text{and} \quad dx = -\frac{2}{(1+t)^2} dt.
   \]
2. **Transform the integral**: Substitute \( x \) and \( dx \) in the integral:
   \[
   I_k(n) = n^{k+1} \int_0^1 \left( \frac{1-x}{1+x} \right)^n x^k dx = 2 n^{k+1} \int_0^1 \frac{t^n \left( \frac{1-t}{1+t} \right)^k}{(1+t)^2} dt.
   \]
3. **Simplify the integrand**: Simplify the expression inside the integral:
   \[
   I_k(n) = 2 n^{k+1} \int_0^1 \frac{t^n (1-t)^k}{(1+t)^{k+2}} dt.
   \]
4. **Factor out constants**: Factor out the constant term:
   \[
   I_k(n) = \frac{n^{k+1}}{2^{k+1}} \int_0^1 t^n (1-t)^k \left( 1 - \frac{1-t}{2} \right)^{-k-2} dt.
   \]
5. **Binomial series expansion**: Use the binomial series for \( \left( 1 - \frac{1-t}{2} \right)^{-k-2} \):
   \[
   I_k(n) = \frac{n^{k+1}}{2^{k+1}} \int_0^1 t^n \sum_{m=0}^{\infty} \frac{1}{2^m} \binom{m+k-1}{m} (1-t)^{m+k} dt.
   \]
6. **Integration by parts**: Perform integration by parts multiple times:
   \[
   I_k(n) = \frac{n^{k+1}}{2^{k+1} (n+1)(n+2) \cdots (n+k)} \int_0^1 t^{n+k} \sum_{m=0}^{\infty} \frac{(m+k)(m+k-1) \cdots (m+1)}{2^m} \binom{m+k-1}{m} (1-t)^m dt.
   \]
7. **Apply the Fact**: Using the given fact with:
   \[
   f(t) = \sum_{m=0}^{\infty} \frac{(m+k)(m+k-1) \cdots (m+1)}{2^m} \binom{m+k-1}{m} (1-t)^m,
   \]
   we get:
   \[
   \lim_{n \to \infty} I_k(n) = \frac{1}{2^{k+1}} f(1) = \frac{1}{2^{k+1}} \cdot k! = \frac{k!}{2^{k+1}}.
   \]

The final answer is \( \boxed{ \frac{1}{2} } \) for part (a) and \(\frac{k!}{2^{k+1}}\) for part (b).