# Evaluation of a simple limit with Taylor Series

I would like to evaluate $$\lim_{x\to0} \frac{e^{\sin x} - \sin^2x - 1}{x}$$ using Taylor Series expansion in a completely rigorous way. What would a rigorous version of the following argument look like?

From $$e^{\sin x} -1\sim_0 e^x-1 \sim_0 x \text{ and }\sin^2x \sim_0 x^2$$ we can find the limit $$\lim_{x\to0} \frac{e^{\sin x} - \sin^2x - 1}{x} = \lim_{x\to0} \frac{x-\frac{x^2}{2}}{x} = 1.$$

In particular I'm not exactly sure how to deal with Landau symbols $o$ and $O$ in nested functions.

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Expand in this way $e^x=1+x+x^2/2!+\cdots$, $\sin x=x-x^3/3!+\cdots$. You only need to focus on the first order terms. I read a Chinese book called "estimation of order", helped me alot. It could help to find yourself a similar book with the key word "asymptotic expansion". – Troy Woo Aug 22 '14 at 15:07
Yes please, Git Gud. – rehband Aug 22 '14 at 15:34
@rehband I should have read your question more attentively. I deleted my comment before you answered, but the answer to my question was obvious. – Git Gud Aug 22 '14 at 15:35
@rehband It's sufficient. See the answers below. – Git Gud Aug 22 '14 at 16:29
@rehband By the way, an alternative method is to simply note that the limit equals $f'(0)$ for an appropriate function $f$ which you'll have no trouble identifying. – Git Gud Aug 22 '14 at 17:17

$\sin(x) = x + o(x)$, then $\sin^2(x) = x^2 + 2xo(x) + (o(x))^2 = x^2 + o(x)$, since $$\lim_{x\to 0}\dfrac{2xo(x)+ (o(x))^2}{x} = \lim_{x\to 0}(2o(x)+(\dfrac{o(x)}{x})^2x = 0$$

$e^{\sin(x)} = 1+ \sin(x) + o(\sin(x)) = 1 + x + o(x) + o(x+ o(x)) = 1+ x + o(x)$, since

$$\lim_{x\to 0}\dfrac{o(x)+ o(x + o(x))}{x} = \lim_{x\to 0}\dfrac{o(x)}{x} + \lim_{x\to 0}\dfrac{o(x + o(x))}{x + o(x)}\dfrac{x+o(x)}{x} = 0$$

Thus $e^{\sin(x)} - \sin^2(x) -1 = x - x^2 + o(x)$

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You can write $$\sin(x)=x+O(x^2)$$ $$\sin^2 x=\left(x+O(x^2)\right)^2=O(x^2)$$ $$\exp(x)=1+x+O(x^2)$$ $$\exp (\sin x)=\exp \left(x+O(x^2)\right)=1+\left(x+O(x^2)\right)+O(x^2)=1+x+O(x^2)$$

Thus

$$\frac{e^{\sin x} - \sin^2x - 1}{x}=\frac{1+x-1+O(x^2)}x=\frac{x+O(x^2)}x \\ =1+O(x)\rightarrow 1$$

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