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Evaluate $$L=\lim_{x \to 0} \frac{e^{\sin(x)}-(1+\sin(x))}{(\arctan(\sin(x)))^2}$$

Method $1$: $$\frac{h^2\left(\frac{e^{h}-1}{h^2}-\frac1{h}\right)}{(\arctan(h))^2}=1^2\left(\frac{1*1}{h}-\frac1{h}\right)=\frac1{h}-\frac1{h}=0$$ Therefore $L=0$.

The identities I have used here to simplify the expression are $$\lim_{x \to 0} \frac{\arctan\left(x\right)}{x}=1$$ $$\lim_{x \to 0} \frac{a^{x}-1}{a}=\ln\left(a\right) \implies \lim_{x \to 0} \frac{e^{x}-1}{x}=\ln\left(e\right)=1$$

Method $2$: $$L=\frac{e^{h}-\left(1+h\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{\left(1+h+\frac{h^{2}}{2!}\right)-\left(1+h\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{\left(\frac{h^{2}}{2!}\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{1}{2}$$ Therefore $L=\frac12$.

I am not at all familiar with $O(n)$ notation BTW.

sato
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  • https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion – lab bhattacharjee Mar 12 '21 at 11:59
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    Method 2 is the correct one. – Crostul Mar 12 '21 at 12:02
  • The mistake appeares to be in Method $1$ in the first equality. Be more precise there. Maybe you can find the mistake on your own. If not, include more steps. 2. Note that $e^h\neq 1+h+h^2/2!$.
    • – vitamin d Mar 12 '21 at 12:03
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    I would be more careful with placing $\lim$ where they belong, at least when writing to present to anyone else. This leads to a lot of apparent "equalities" which are not actually equal. – aschepler Mar 12 '21 at 12:24
  • @Crostul I really wonder how you can say that "method 2" (which isn't a method at all as presented) is the "correct one"? – trancelocation Mar 12 '21 at 14:21
  • @trancelocation it's almost correct, a part from a $+o(h^2)$. Come on, I would say it's ok. – Crostul Mar 12 '21 at 15:19