Did you mean for this to be a calculus problem? Calculus can be used to prove the binomial expansion (for #|x|<1#) #(1+x)^(-1/2)=1-\frac{1}{2}x+\frac{-1/2\cdot -3/2}{2!}x^{2}+\frac{-1/2\cdot -3/2\cdot -5/2}{3!}x^3+\frac{-1/2\cdot -3/2\cdot -5/2\cdot -7/2}{4!}x^{4}+\cdots#.
This simplifies to
#(1+x)^{-1/2}=1-x/2+\frac{3}{8}x^{2}-\frac{5}{16}x^{3}+\frac{35}{128}x^{4}+\cdots# for #|x|<1#.
This means #\frac{1}{\sqrt{.99}}=(1+(-.01))^{-1/2}=1+.01/2+\frac{3}{8}(.01)^2+\frac{5}{16}(.01)^{3}+\frac{35}{128}(.01)^{4}+\cdots#
#\approx 1+0.005+0.0000375+0.0000003125+0.00000000273438\approx 1.00504#
