What is stopping criteria for Newtons Method? 
Use newtons method to find solutions accurate to within $10^{-4}$ for the following: $$\\x^3-2x^2-5=0,\qquad[1,4]$$

Using : $p_{0}=2.0$ 
$\Rightarrow $ 
My question for the newtons method is what is the stopping criteria for it? How does one derive the function value? Does one use $$\\ \frac{|P_{n}-P_{n-1}|}{|P_n|}$$ or $$|P_{n}-P_{n-1}|$$
 A: You are searching for the roots of a polynomial $p$. In this case it is possible to compute a running error bound for $p$, i.e. a number $\mu$ such that the computed value $\hat{y}$ of $y = p(x)$ satisfies $$|y - \hat{y}| \leq \mu u,$$ where $u$ is the unit round off error. When the error bound exceeds the absolute value of $\hat{y}$ there is no point in doing additional iterations.
The running error bound depends on the algorithm used to evaluate $p(x) = \sum_{i=0}^n a_i x^i$. Horner's method computes $$p_0 = a_n, \quad p_i = x p_{i-1} + a_{n-i}, \quad i = 1,2,\dotsc,n.$$ If $a_i$ and $x$ are machine numbers, then a running error bound can be computed concurrently as follows 
$$\mu_0 = 0, \quad z_j = p_{j-1} x, \quad p_j = z_j + a_{n-j}, \quad \mu_j = \mu_{j-1} |x| + |z_j| + |p_j|, \quad j=1,2,\dotsc, n.$$
The algorithm returns $y = p_n$ and $\mu = \mu_n$. It is possible, to reduce the cost of this algorithm from 5n flops to 4n flops, but this is not a critical issue at this juncture.
In the general case where your function is not a polynomial you must proceed with caution. 
Ideally, you merge a rapidly convergent routine such as Newton's method or the secant method with the bisection method and maintain a bracket around the root. If you trust the sign of the computed values of $y = f(x)$, then you have a robust way of estimating the error. Here a running error bound or interval arithmetic can be helpful.
I must caution you against using the correction $$\Delta x \approx - f'(x)/f(x)$$ as an error estimate. If $0 = f(x + \Delta x)$, then by Taylor's theorem
$$0 = f(x) + f'(x) \Delta x + \frac{1}{2} f''(\xi_x) (\Delta x)^2 $$ for some $\xi_x$ between $x$ and $x + \Delta x$. It follows, that $\Delta x \approx - f'(x)/f(x)$ is a good approximation only when the second order term can be ignored. This is frequently the case, but I would not count on it in general.
Occasionally, it is helpful to remember that Newton's method exhibits one sided convergence in the limit, i.e. if the root is a simple, then the residuals $f(x_n)$ eventually have the same sign. Deviation from this behavior indicates that you are pushing against the limitations imposed by floating point arithmetic.
EDIT: Newton's and other iterative methods are frequently terminated when either  $$|f(x_n)| < \epsilon$$ or $$|x_n - x_{n-1}| < \delta$$
where $\epsilon$ and $\delta$ are user-defined thresholds. This is a dangerous practice which is unsuited for general purpose software. A specific example is the equation $$f(x) = 0,$$
where $f(x) = \exp(-\lambda x)$ and $\lambda > 0$. Obviously, there are no solutions, but this is not the point. Newton's method yields $$x_{n+1} = x_n + \frac{1}{\lambda} \rightarrow \infty, \quad n \rightarrow \infty$$
It follows that the residual will eventually drop below the user's threshold. Moreover, if $\lambda$ is large enough, then the routine will immediately exit "succesfully", because $x_{n+1} - x_n$ is small enough.
Writing a robust nonlinear solver is a nontrivial exercise. You have to maintain a bracket around the root. You must be able to trust at least the computed signs. To that end, the user must supply not only a function which computes $f$, but also a reliable error bound.  
A: Without computing higher derivatives the best estimate for the momentaneous error $\xi-x_n$ is the size of the next correction, namely $-{f(x_n)\over f'(x_n)}$. After all, if there would be a better estimate of the error using the available information, one would use this better estimate instead for the correction. In numerical praxis this means that one stops after the first correction which is smaller than the desired accuracy.
