If  $f\left(x\right)=\underset{n\to \infty }{\lim }\frac{x^n-x^{-n}}{x^n+x^{-n}},x>1$ and  $\int f\left(x\right){\tan }^{-1}xdx=x.g\left(x\right)-\frac{1}{2}\ln (1+x^2)+c$ , then number of solution for the equation  $g^{-1}\left(x\right)=x$ is:

We begin by evaluating the given function:

f(x) = lim (n → ∞) (x^n - x^(-n)) / (x^n + x^(-n)), where x > 1.    

To simplify, divide both numerator and denominator by x^n:

f(x) = lim (n → ∞) (1 - x^(-2n)) / (1 + x^(-2n)).    

As x > 1, x^(-2n) approaches 0 as n → ∞. Thus:

f(x) = (1 - 0) / (1 + 0) = 1.    

Integral Evaluation

The problem provides the integral:

∫ f(x) tan^(-1)(x) dx = x * g(x) - (1/2) ln(1 + x^2) + C.    

Since f(x) = 1, the integral simplifies as:

∫ tan^(-1)(x) dx = x * tan^(-1)(x) - (1/2) ln(1 + x^2) + C.    

Therefore, the function g(x) is:

g(x) = tan^(-1)(x), for x > 1.    

Domain of g(x) and g^(-1)(x)

For x > 1, tan^(-1)(x) is within the range:

g(x) ∈ (π/4, π/2).    

Thus, the domain of the inverse function g^(-1)(x) is:

Domain of g^(-1)(x) = (π/4, π/2).    

Since g^(-1)(x) = tan(x), we analyze the equation:

tan(x) = x, within the interval (π/4, π/2).    

Solution Analysis

The function tan(x) is strictly increasing in the interval (π/4, π/2), while x is a linear function. Given the steep growth of tan(x) in this range, it does not intersect with x. Therefore, the equation:

g^(-1)(x) = x, i.e., tan(x) = x,    

has no solution in the interval (π/4, π/2).

Conclusion

Using the above analysis, the number of solutions for the equation g^(-1)(x) = x is: Zero.