Let $\alpha$ be a solution satisfying the equation $|x|=e^{-x}.$ Let $I_n=\int_0^{\alpha} (xe^{-nx}+\alpha x^{n-1})dx\ (n=1,\ 2,\ \cdots).$
Find $\lim_{n\to\infty} n^2I_n.$
1. **Finding the value of $\alpha$:**

   The equation given is \( |x| = e^{-x} \). Since \( x \) must be non-negative (as \( e^{-x} \) is always positive), we can write:
   \[
   x = e^{-x}
   \]
   To find the solution, we can solve this equation graphically or numerically. From the graphs of \( y = x \) and \( y = e^{-x} \), we observe that they intersect at a point between 0.5 and 0.6. For the sake of this problem, we denote this intersection point as \( \alpha \).

2. **Evaluating the integral \( I_n \):**

   The integral is given by:
   \[
   I_n = \int_0^{\alpha} \left( x e^{-nx} + \alpha x^{n-1} \right) \, dx
   \]

   We split this into two integrals:
   \[
   I_n = \int_0^{\alpha} x e^{-nx} \, dx + \int_0^{\alpha} \alpha x^{n-1} \, dx
   \]

3. **Evaluating the first integral using integration by parts:**

   Let \( u = x \) and \( dv = e^{-nx} \, dx \). Then \( du = dx \) and \( v = -\frac{1}{n} e^{-nx} \). Using integration by parts:
   \[
   \int x e^{-nx} \, dx = -\frac{x}{n} e^{-nx} \bigg|_0^{\alpha} + \frac{1}{n} \int_0^{\alpha} e^{-nx} \, dx
   \]
   Evaluating the boundary terms:
   \[
   -\frac{\alpha}{n} e^{-n\alpha} + \frac{1}{n} \left[ -\frac{1}{n} e^{-nx} \right]_0^{\alpha} = -\frac{\alpha}{n} e^{-n\alpha} - \frac{1}{n^2} \left( e^{-n\alpha} - 1 \right)
   \]
   Simplifying:
   \[
   \int_0^{\alpha} x e^{-nx} \, dx = -\frac{\alpha}{n} e^{-n\alpha} - \frac{e^{-n\alpha} - 1}{n^2}
   \]

4. **Evaluating the second integral:**

   \[
   \int_0^{\alpha} \alpha x^{n-1} \, dx = \alpha \int_0^{\alpha} x^{n-1} \, dx
   \]
   Using the power rule for integration:
   \[
   \alpha \left[ \frac{x^n}{n} \right]_0^{\alpha} = \frac{\alpha^n}{n}
   \]

5. **Combining the results:**

   \[
   I_n = -\frac{\alpha}{n} e^{-n\alpha} - \frac{e^{-n\alpha} - 1}{n^2} + \frac{\alpha^n}{n}
   \]

6. **Multiplying by \( n^2 \) and taking the limit:**

   \[
   n^2 I_n = n^2 \left( -\frac{\alpha}{n} e^{-n\alpha} - \frac{e^{-n\alpha} - 1}{n^2} + \frac{\alpha^n}{n} \right)
   \]
   Simplifying:
   \[
   n^2 I_n = -n \alpha e^{-n\alpha} - (e^{-n\alpha} - 1) + n \alpha^n
   \]
   As \( n \to \infty \):
   - \( n \alpha e^{-n\alpha} \to 0 \) because \( e^{-n\alpha} \) decays faster than \( n \) grows.
   - \( e^{-n\alpha} \to 0 \).
   - \( n \alpha^n \to 0 \) because \( \alpha < 1 \).

   Therefore:
   \[
   \lim_{n \to \infty} n^2 I_n = 1
   \]

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