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=) 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.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In 11e) Can you just please explain how radical 3 x was factored and how it got replaced with other stuff
- f4 is infinitesimal of order 2 and its principal part is x →0. f5(x)=√x + x² - √x + x² ~ 11e) Even though it is possible to notice that √x + x² 3x (x0), 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 11f) As x → 0, it results that fo(x) = = = = 17x², 1 √((1+z) ¹/3 − 1) + x² = - 7 ( 1² 2 + 0(2)) - 24/3 with respect to x, as f5 is infinitesimal of order 4/3 and its principal part is 1 3 as x → 0. o(x)) + x² = +0(x²/³), as x → 0. x4/3 " with respect to x, ee* - ecosx = e¹+x+o(x) - el-x² +0(x²) = =e p (p²+0(x) - e- x² +0(x²))
Transcribed Image Text:- f4 is infinitesimal of order 2 and its principal part is x →0. f5(x)=√x + x² - √x + x² ~ 11e) Even though it is possible to notice that √x + x² 3x (x0), 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 11f) As x → 0, it results that fo(x) = = = = 17x², 1 √((1+z) ¹/3 − 1) + x² = - 7 ( 1² 2 + 0(2)) - 24/3 with respect to x, as f5 is infinitesimal of order 4/3 and its principal part is 1 3 as x → 0. o(x)) + x² = +0(x²/³), as x → 0. x4/3 " with respect to x, ee* - ecosx = e¹+x+o(x) - el-x² +0(x²) = =e p (p²+0(x) - e- x² +0(x²))
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