`Let f(x) and g(x)` be two continuous function and `h(x)= lim_(n to oo) (x^(2n).f(x)+x^(2m).g(x))/((x^(2n)+1)).` if the limit of h(x) exists at x=1, then one root of f(x)-g(x) =0 is _____.

Answer

To solve the problem, we need to analyze the limit of the function \( h(x) \) as \( n \) approaches infinity and determine the conditions under which it exists at \( x = 1 \).

### Step-by-Step Solution:

1. **Define the function \( h(x) \)**:
\[ h(x) = \lim_{n \to \infty} \frac{x^{2n} f(x) + x^{2m} g(x)}{x^{2n} + 1} \]

2. **Consider the limit as \( x \) approaches 1 from the right**:
\[ \lim_{x \to 1^+} h(x) = \lim_{x \to 1^+} \lim_{n \to \infty} \frac{x^{2n} f(x) + x^{2m} g(x)}{x^{2n} + 1} \]
Here, as \( n \to \infty \) and \( x > 1 \), \( x^{2n} \) dominates over 1 in the denominator.

3. **Simplify the expression**:
\[ h(x) = \lim_{n \to \infty} \frac{x^{2n} f(x)}{x^{2n}} + \lim_{n \to \infty} \frac{x^{2m} g(x)}{x^{2n}} = f(x) + \lim_{n \to \infty} \frac{x^{2m} g(x)}{x^{2n}} \]
Since \( x^{2n} \to \infty \), the second term goes to 0:
\[ h(x) = f(x) \quad \text{for } x > 1 \]

4. **Now consider the limit as \( x \) approaches 1 from the left**:
\[ \lim_{x \to 1^-} h(x) = \lim_{x \to 1^-} \lim_{n \to \infty} \frac{x^{2n} f(x) + x^{2m} g(x)}{x^{2n} + 1} \]
Here, as \( n \to \infty \) and \( x < 1 \), \( x^{2n} \) approaches 0.

5. **Simplify this expression**:
\[ h(x) = \lim_{n \to \infty} \frac{0 \cdot f(x) + x^{2m} g(x)}{1} = x^{2m} g(x) \]
Therefore, as \( x \to 1^- \):
\[ h(x) = g(x) \quad \text{for } x < 1 \]

6. **Set the limits equal to ensure continuity**:
Since \( h(x) \) must exist at \( x = 1 \), we equate the right-hand limit and left-hand limit:
\[ f(1) = g(1) \]

7. **Conclusion**:
The condition \( f(1) = g(1) \) implies that \( x = 1 \) is a root of the equation \( f(x) - g(x) = 0 \).

Thus, one root of \( f(x) - g(x) = 0 \) is:
\[ \boxed{1} \]