Let $ I_n\equal{}\int_0^{\sqrt{3}} \frac{1}{1\plus{}x^{n}}\ dx\ (n\equal{}1,\ 2,\ \cdots)$.

(1) Find $ I_1,\ I_2$.


(2) Find $ \lim_{n\to\infty} I_n$.
1. **Finding \( I_1 \) and \( I_2 \):**

   For \( I_1 \):
   \[
   I_1 = \int_0^{\sqrt{3}} \frac{1}{1 + x} \, dx
   \]
   This is a standard integral:
   \[
   I_1 = \left[ \ln(1 + x) \right]_0^{\sqrt{3}} = \ln(1 + \sqrt{3}) - \ln(1 + 0) = \ln(1 + \sqrt{3}) - \ln(1) = \ln(1 + \sqrt{3})
   \]

   For \( I_2 \):
   \[
   I_2 = \int_0^{\sqrt{3}} \frac{1}{1 + x^2} \, dx
   \]
   This is also a standard integral:
   \[
   I_2 = \left[ \arctan(x) \right]_0^{\sqrt{3}} = \arctan(\sqrt{3}) - \arctan(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3}
   \]

2. **Finding \( \lim_{n \to \infty} I_n \):**

   We need to evaluate:
   \[
   \lim_{n \to \infty} \int_0^{\sqrt{3}} \frac{1}{1 + x^n} \, dx
   \]

   We split the integral into two parts:
   \[
   \int_0^{\sqrt{3}} \frac{1}{1 + x^n} \, dx = \int_0^1 \frac{1}{1 + x^n} \, dx + \int_1^{\sqrt{3}} \frac{1}{1 + x^n} \, dx
   \]

   For the first part:
   \[
   \int_0^1 \frac{1}{1 + x^n} \, dx
   \]
   As \( n \to \infty \), \( x^n \) approaches 0 for \( 0 \leq x < 1 \). Thus, \( \frac{1}{1 + x^n} \) approaches 1. Therefore:
   \[
   \lim_{n \to \infty} \int_0^1 \frac{1}{1 + x^n} \, dx = \int_0^1 1 \, dx = 1
   \]

   For the second part:
   \[
   \int_1^{\sqrt{3}} \frac{1}{1 + x^n} \, dx
   \]
   As \( n \to \infty \), \( x^n \) becomes very large for \( x > 1 \). Thus, \( \frac{1}{1 + x^n} \) approaches 0. Therefore:
   \[
   \lim_{n \to \infty} \int_1^{\sqrt{3}} \frac{1}{1 + x^n} \, dx = 0
   \]

   Combining these results:
   \[
   \lim_{n \to \infty} \int_0^{\sqrt{3}} \frac{1}{1 + x^n} \, dx = \lim_{n \to \infty} \int_0^1 \frac{1}{1 + x^n} \, dx + \lim_{n \to \infty} \int_1^{\sqrt{3}} \frac{1}{1 + x^n} \, dx = 1 + 0 = 1
   \]

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