• Trivial cases, where limit as x® a, we just put x = a with no problem.
• Cases which involve division by zero, in which case factorise with the top and then take the limit, for example limit as x to 2 of:
  

  (x-2)/(x2-x-2).

  Check with you calculator, do get 1/3.

Using this we find that xⁿ goes to nx^n-1. For example If f(x) = x² then

lim h® 0  f(x+h)-f(x) h = lim h® 0  (x+h)2-x2 h

= lim h® 0  x2+2xh+h2-x2 h = lim h® 0  2xh+h2 h

= lim h® 0  2x+h = 2x.

• Algorithm to find roots of polynomials
• Will use the fact that the derivative is the slope and that the derivative of xⁿ is nx^n-1.
• We begin by making a guess for root of f(x), which in general not be right. Will have f(x₀). Now look at the slope of the function at x₀. If m then would guess that f hits zero at
  

  x1 = x0-f(x0)/m,

  which would the case if the f was linear. This 2nd guess will also not normally be a root but if we keep going we will find one. So look at f(x₁) and the slope at x₁, use these to get x₂ and so on!
• To find a root of x³-x²+x-1 = 0 we need to make an initial guess. So I chose x₀ = 0. f(x) = -1 so this is not a root. Now using the algorithm we can determine
  

  x1 = x0-f(x0)/m.

  x₀ = -1 and f(0) = -1 but we need to determine m, the slope. The slope at x₀ is given by the derivative at x₀ which is
  

  x3-x2+x-1 ® 3x2-2x+1.

  The slope at x₀ then is
  

  3(0)2-2(0)+1 = 1

  and
  

  x1 = 0- (-1)/1 = 1.

  Then x₁ = 1 and f(1) = 1-1+1-1 = 0 so we have found the root already.
• Normally will not get an exact solution and we look for x where | f(x) | < 0.001 to get close to a root.
• We only get a root this way. In order to get all of them would divide the polynomial by (x-x_root) to get a simpler polynomial which we would again apply the algorithm too. Depending on our initial guess we converge to different roots.