Newton's method seeks to approximate a solution $f(x)=0$ that starts with an initial approximation $x_{0}$ and successively defines a sequence $x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}.$ For the given choice of $f$ and $x_{0},$ write out the formula for $x_{n+1}$ . If the sequence appears to converge, give an exact formula for the solution $x,$ then identify the limit $x$ accurate to four decimal places and the smallest $n$ such that $x_{n}$ agrees with $x$ up to four decimal places.
$$f(x)=e^{x}-2, \quad x_{0}=1$$