Algorithm to find the square root of a number

So you always used sqrt() function to find out the square roots of a number .. So what if i ask you to write a function that will display the square root without using the math library functions.. STUNNED ?
       Well lets try to figure out what we do in finding the square root of a number.So how do we start.The first thing that comes into my mind is that square root can be any real value less than the given number.Hence there are infinite possibilities and we cannot loop through to check if multiplying the number to itself will give us the number.That would be a never ending process and we do not know the increment value.So what do we do now ?
       Ok.What if we take any number as the initial guess and start.When we find the square of the number is greater than the number than we know we have to select a number that is less than the current and if it is smaller than choose a greater number.In this way after a fixed number of iterations we will eventually reach near to the square root.This method is also called as Approximation by Bisection.

def sqrt(x,epsilon):
    assert((x>0)&&(epsilon>0)),"Initial Conditions not met"
    low=0
    high=x
    val=(low+high)/2
    count=1
    while val**2-x>epsilon and crt<100:
        crt+=1
        if(val**2>x)
            high=val
        else:
            low=val
        val=(high+low)/2
   return val

However even this cannot find to much approximation and what if in count number of times it doesn't reach the required value.For this we have another method called the Newtons method where we use the initial guess to make a better guess.It works like taking 1 as the initial guess.Then

1+(2/1)=3  3/2=1.5
1.5+(2/1.5)=2.83  2.83/2=1.416 ~ 1.414 sqrt of 2

Hence using Newtons Method of Approximation we get:

float sqrt(float x)
{
 float ng;
 float g=x/2;
 float d=(g*g)-x;
 d=(d>0)?d:-d;
 while(d>EPS)
 {
  ng=(g+(x/g))*0.5;
  g=ng;
  d=(g*g)-x;
  d=(d>0)?d:-d;
  //cout<<g<<endl;
 }
 return g;
}