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I am looking for help with this question

By looking at the Taylor series, decide which of the following functions $\ln(1+y^2) \; \; \; \; \; \; \; \; \sin(y^2)\; \; \; \; \; \; \; \;1-\cos(y)$ is largest and which is smallest for values of $y$ near $0$.

I've done the Taylor expansions about $0$. I've written them below with the first few terms of the expansions.

$f(x)=\ln(1+x^2)$ $=x^2- \frac {x^4}{2}+\frac {x^6}{3}-\frac {x^8}{4}+\frac{x^{10}}{5}$

$--$

$f(x)=\sin(x^2)$ $=x^2- \frac {x^6}{6}+\frac {x^{10}}{120}-\frac {x^{14}}{5040}+\frac{x^{18}}{362880}$

$--$

$f(x)=1-\cos(x)$ $=\frac {x^2}{2}- \frac {x^4}{24}+\frac {x^6}{720}-\frac {x^8}{40320}+\frac{x^{10}}{3628800}$

How do I find which function is largest/smallest for values of $x$ near $0$? Each of these functions tends toward the point of origin. I'm having troubles grasping the intuition of the question.

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    You could always cheat and use a calculator to evaluate each of the functions at, say, $x=.001$ to see how they compare.2012-03-29

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They are all equall to zero for the exact argument 0, but if you are not exactly at the origin but only close, say at $\epsilon$, then $\epsilon^2>\epsilon^2/2$. Similarly if the first term is equal, then the next one decides which one is larger for small arguments.

When the powers are not identical, then (for the limit $x\rightarrow 0$) the lower power dominates the other one ($x^2 < n x^4$ for small enough $x$. the larger n is, the smaller x has to be for this to be true, but it is true nontheless when you are "near" 0).

Reading up on big O notation might help if you want my last sentence in a more mathematical formulation http://en.wikipedia.org/wiki/O_notation .