(a) Show that for every $n\in\mathbb{N}$ there is exactly one $x\in\mathbb{R}^+$ so that $x^n+x^{n+1}=1$. Call this $x_n$.
(b) Find $\lim\limits_{n\rightarrow+\infty}x_n$.
### Part (a)
1. Consider the function \( f(x) = x^n + x^{n+1} \) for \( x \in \mathbb{R}^+ \).
2. The function \( f(x) \) is continuous and strictly increasing on \( \mathbb{R}^+ \). This is because both \( x^n \) and \( x^{n+1} \) are continuous and strictly increasing functions for \( x > 0 \).
3. As \( x \) approaches 0, \( f(x) \) approaches 0. As \( x \) approaches \( +\infty \), \( f(x) \) approaches \( +\infty \).
4. By the Intermediate Value Theorem, since \( f(x) \) is continuous and strictly increasing, for any \( y \) in the interval \( (0, +\infty) \), there exists a unique \( x \in \mathbb{R}^+ \) such that \( f(x) = y \).
5. Specifically, for \( y = 1 \), there exists a unique \( x_n \in \mathbb{R}^+ \) such that \( x_n^n + x_n^{n+1} = 1 \).

Thus, we have shown that for every \( n \in \mathbb{N} \), there is exactly one \( x \in \mathbb{R}^+ \) such that \( x^n + x^{n+1} = 1 \). We denote this unique \( x \) as \( x_n \).

\(\blacksquare\)

### Part (b)
1. We need to find \( \lim_{n \to +\infty} x_n \).
2. First, observe that \( 0 < x_n < 1 \). This is because if \( x_n \geq 1 \), then \( x_n^n + x_n^{n+1} \geq 1 + x_n > 1 \), which contradicts \( x_n^n + x_n^{n+1} = 1 \).
3. Assume \( x_{n+1} < x_n \). Then \( 1 = x_n^n + x_n^{n+1} > x_{n+1}^{n+1} + x_{n+1}^{n+2} \). This is a contradiction because \( x_{n+1}^{n+1} + x_{n+1}^{n+2} \) should also equal 1.
4. Therefore, \( x_{n+1} \geq x_n \), meaning \( \{x_n\} \) is an increasing sequence.
5. Since \( \{x_n\} \) is increasing and bounded above by 1, it converges to some limit \( \alpha \) where \( 0 < \alpha \leq 1 \).
6. Suppose \( 0 < \alpha < 1 \). Then \( \lim_{n \to +\infty} \ln x_n^n = \lim_{n \to +\infty} n \ln x_n = -\infty \). Hence, \( \lim_{n \to +\infty} x_n^n = 0 \).
7. Since \( 0 < x_n^{n+1} < x_n^n \), by the Squeeze Theorem, \( \lim_{n \to +\infty} x_n^{n+1} = 0 \).
8. Thus, \( \lim_{n \to +\infty} (x_n^n + x_n^{n+1}) = 0 \), which contradicts \( x_n^n + x_n^{n+1} = 1 \).
9. Therefore, \( \alpha = 1 \).

\(\blacksquare\)

The final answer is \( \boxed{ \lim_{n \to +\infty} x_n = 1 } \).