If `f(X)` is a polynomial satisfying `f(x)f(1/x)=f(x)+f(1/x)` and `f(2)gt1`, then `lim_(x rarr 1)f(x)` is

Answer

To solve the problem, we need to find the limit \(\lim_{x \to 1} f(x)\) given the polynomial function \(f(x)\) that satisfies the equation:

\[ f(x)f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) \]

and the condition \(f(2) > 1\).

### Step-by-Step Solution:

1. **Understanding the Equation**:
We start with the equation:
\[ f(x)f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) \]
This implies a relationship between \(f(x)\) and \(f\left(\frac{1}{x}\right)\).

2. **Rearranging the Equation**:
Rearranging gives us:
\[ f(x)f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) = 0 \]
This can be factored as:
\[ (f(x) - 1)(f\left(\frac{1}{x}\right) - 1) = 1 \]

3. **Defining a New Function**:
Let \(g(x) = f(x) - 1\). Then we can rewrite our equation as:
\[ g(x)g\left(\frac{1}{x}\right) = 1 \]
This indicates that \(g(x)\) and \(g\left(\frac{1}{x}\right)\) are multiplicative inverses.

4. **Analyzing the Function**:
The equation \(g(x)g\left(\frac{1}{x}\right) = 1\) suggests that \(g(x)\) could be of the form \(g(x) = x^n\) for some integer \(n\). This leads to:
\[ f(x) = g(x) + 1 = x^n + 1 \]

5. **Using the Condition \(f(2) > 1\)**:
We know \(f(2) > 1\):
\[ f(2) = 2^n + 1 > 1 \implies 2^n > 0 \]
This is true for all integers \(n\).

6. **Finding the Limit**:
Now we need to find:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} (x^n + 1) = 1^n + 1 = 1 + 1 = 2 \]

### Conclusion:
Thus, the limit is:
\[ \lim_{x \to 1} f(x) = 2 \]