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I tried to do this like that:
$$ \sin(x^2) = \mathcal{o}(x) \iff \lim_{x \to 0} \frac{\sin(x^2)}{x} = 0$$
we could get $\sin(x^2)$ from Taylor series.
For $x_0 = 0$, $T_n = 0$ for every $n$.
So from Peano remainder ($\lim_{x\to 0}r(x)=0$) we have:
$$\lim_{x\to 0}\sin(x^2) = 0$$
which means $$\lim_{x \to 0} \frac{\sin(x^2)}{x} = 0$$

Is this correct?

Salech Alhasov
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Badziubel
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2 Answers2

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Why not use that $$\lim_{x\to 0}\frac{\sin x^2}{x^2}=1 ?$$ Note then that $$\lim_{x\to 0}\frac{\sin x^2}{x}=\lim_{x\to 0}\frac{\sin x^2}{x^2}x=1\cdot 0=0 $$

Pedro
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  • Thanks for answering :)I feel like it's a stupid question but how do we know that and how does it help us? :( – Badziubel Sep 02 '13 at 21:17
  • @Badziubel The limit in the first line substitutes $u=x^2$ into the limit identity $\lim_{u \rightarrow 0} (\sin u)/u = 1$ which can be proven in a variety of ways (see http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1); the second line arises from the result that "the limit of the product is equal to the product of the limits". – Andrew D Sep 02 '13 at 22:08
  • @Badziubel I have added something to clarify. – Pedro Sep 03 '13 at 02:22
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You can use L'Hopital's rule if you want to show $\lim_{x \to 0} \frac{\sin x^2}{x} = 0$, because taking derivatives of numerator and denominator gives $\lim_{x \to 0} \frac{2x \cos x^2}{1} = 0$.

user2566092
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