There's a few different ways this can be done. Probably the most straight-forward approach is to let #f(x)=(4-x)^(1/2)# and write the Taylor series about #x=0# as #f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots#
Since #f(x)=(4-x)^(1/2)#, #f(0)=4^{1/2}=2#, #f'(x)=\frac{1}{2}(4-x)^{-1/2}*(-1)#, #f'(0)=-\frac{1}{2}*4^{-1/2}=-\frac{1}{4}#,
#f''(x)=\frac{1}{4}(4-x)^{-3/2}*(-1)#, #f''(0)=-\frac{1}{4}*4^{-3/2}=-\frac{1}{32}#, etc...
Eventually this leads to the Taylor series #2-\frac{1}{4}x-\frac{1}{64}x^2-\frac{1}{512}x^{3}-\cdots#, which converges to #f(x)=(4-x)^{1/2}# for #|x|<4#.
Another approach is to use a bit of slick algebra along with the generalized binomial theorem (discovered by Newton) that #(1+x)^{p}=1+px+\frac{p(p-1}}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots# for #|x|<1#.
We have: #(4-x)^{1/2}=4^{1/2}(1-x/4)^{1/2}=2(1+(-x/4))^{1/2}#. Now use the generalized binomial theorem with #-x/4# in place of #x# and #p=1/2# to get, for #|-x/4|<1 \Leftrightarrow |x|<4#
#(4-x)^{1/2}#
#=2(1+(-x/4))^{1/2}#
#=2(1+(1/2)(-x/4)+\frac{1/2*-1/2}{2!}(-x/4)^2+\frac{1/2*-1/2*-3/2}{3!}(-x/4)^3+\cdots)#
#=2-x/4-\frac{1}{64}x^2-\frac{1}{512}x^{3}-\cdots#
