I am trying to understand example 3.4.5 in John and Barbara Hubbard's second edition of Vector Calculus, Linear Algebra, and Differential Forms. It provides the taylor expansion of $sin(x+y^2)$ by using the chain rule for Taylor polynomials. The result is given by substituting $x+y^2$ into the expansion $sin(u)=u-u^3/6+o(u^3)$, which I can follow. But when they plug in the error term becomes $o(x^2+y^2)^{\frac 32}$ instead of $o(x+y^2)^3$. I am struggling to prove rigorously that the latter implies the former. Any hints please?
1 Answer
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We have $$ \frac{(x+y^2)^2}{x^2+y^2} = \frac{x^2+y^2(2x+y^2)}{x^2+y^2}, $$ so when $(x,y)\rightarrow(0,0)$ we have $$ \left|\frac{(x+y^2)^2}{x^2+y^2}\right|\le1, $$ from which it follows that we can replace $o\left((x+y^2)^3\right)$ with $o\left((x^2+y^2)^\frac{3}{2}\right)$.
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$\begingroup$ You mean that the limit (or lim sup) $=1$, correct? $\endgroup$ Sep 27, 2015 at 2:14