Let #f(x)=sqrt{x}=x^{1/2}# so that #f'(x)= 1/2 x^{-1/2}#, #f''(x)=-1/4 x^{-3/2}#, and #f'''(x)=3/8 x^{-5/2}#.
Since #f(1)=1#, #f'(1)=1/2#, #f''(1)=-1/4#, and #f'''(1)=3/8#, the Taylor polynomial of degree #n=3# for #f# centered at #a=1# is
#P_{3}(x)=1+1/2 (x-1) - (1/4)/(2!) (x-1)^{2}+ (3/8)/(3!) (x-1)^{3}#
#=1+1/2 (x-1) - 1/8 (x-1)^{2} + 1/16 (x-1)^{3}#.
If we let #E_{3}(x)=f(x)-P_{3}(x)# be the error in using #P_{3}(x)# to approximate #f(x)# for #x approx a = 1#, then Taylor's Theorem guarantees
#|E_{3}(x)|=|E_{n}(x)|\leq M/((n+1)!) * |x-a|^{n+1}=M/(4!)|x-1|^{4}#, where #|f''''(t)|\leq M# for all #t# between #x# and #a=1#.
Since #f''''(x)=-15/16 x^{-7/2}#, we have #|f''''(t)| = 15/16 t^{-7/2}#, which is maximized for #t\in [0.9,1.1]# at #t=0.9#. Therefore, we can take #M=|f''''(0.9)|=15/16 * 0.9^{-7/2}approx 1.35557#.
Hence, rounding up gives #|E_{3}(x)|\leq 1.3556/24 * 0.1^{4}approx 0.00000565# for #x in [0.9,1.1]#.
You can double-check this by noting that #sqrt{.9}-P_{3}(.9) approx -0.00000420# and #sqrt{1.1}-P_{3}(1.1) approx -0.00000365#.
