=) Even though it is possible to notice that √x + x² ~ 3x (x → 0), this does not allow to conclude that Vx+x²-3x + x² is equivalent to x², as x → 0. Indeed, recalling that (1+x)a - 1 = ax + o(x)~ ax, as x→ 0, it results that f5(x)=√x+x²-3x+x² √((1+x)¹/3-1) + x² = √z ( 3 x + 0(x)) - - = = + x² = 1 24/3 + 0(2¹/3), as z →0. x 1 fs is infinitesimal of order 4/3 and its principal part is ¹/3, with respect to x, as x→0.
=) Even though it is possible to notice that √x + x² ~ 3x (x → 0), this does not allow to conclude that Vx+x²-3x + x² is equivalent to x², as x → 0. Indeed, recalling that (1+x)a - 1 = ax + o(x)~ ax, as x→ 0, it results that f5(x)=√x+x²-3x+x² √((1+x)¹/3-1) + x² = √z ( 3 x + 0(x)) - - = = + x² = 1 24/3 + 0(2¹/3), as z →0. x 1 fs is infinitesimal of order 4/3 and its principal part is ¹/3, with respect to x, as x→0.
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter8: Polynomials
Section8.4: Zeros Of A Polynomial
Problem 22E
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Can you just please explain how radical 3 x was factored and how it got replaced with other stuff
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