For any function #f(x)# the taylor expansion of that function about a variable #a# will be:

#f(x) = f(a) + f'(a)(x-a) +f''(a) (x-a)^2 /(2!) +...#

#f(x) = sum_(n=0)^infty f^n(a) (x-a)^n / (n!)#

The function #sqrt(x)# presents us with an easily exploitable property. All of its derivatives, evaluated at #x = 1# are also 1:

#f(x) = sqrt(x) = x^(1/2)#

#f'(x) = x^(-1/2)#

#f''(x) = x^(-3/2)#

#f'''(x) = x^(-5/2)#

It should be immediately obvious that #1^n = 1# for any #n#.

Thus, we can replace all #f^n(a)# terms with 1. From that point forward, writing the solution is easy:

#f(x) = 1 + (x-1) + (x-1)^2 / (2!) +...#

#f(x) = sum_(n=0)^infty (x-1)^n /(n!)#