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 Dec 31 comment Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$ It was only the way I wrote things that seemed not formal enough to me. Looking at the answer to this post, it looks like that's only my perception that was wrong. I did not start with the definition of $\tan(x) = \frac{\sin(x)}{\cos(x)}$ but, if I am not mistaken, I think this definition is based on the exact thing that makes me uneasy in the way I wrote the first steps of the limit above Dec 31 accepted Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$ Dec 31 awarded Scholar Dec 31 comment Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$ Looks indeed better to me. Isn't this definition based on the "opp/hyp/adj" definition though ? Dec 30 asked Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$ Aug 8 awarded Tumbleweed May 19 awarded Supporter Apr 22 answered Finding ALL solutions to $2(\sin^2(x)) - 5\sin(x)-3 = 0$? Apr 14 awarded Teacher Dec 15 comment Autocorrelation derivation using fourier transform Oh my god yeah I was tired when I answered you. $R_{x}(\tau) = \frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}{cos(\omega{t})cos(\omega(t-\tau))dt‌​}$. Anyway, it doesn't change the result of the auto-correlation. However, My question is more about the fourier derivation of the auto-correlation, no matter how much I could be wrong/inexact with the integral derivation. I would be pleased if you could comment on that in priority ! Dec 15 comment Autocorrelation derivation using fourier transform Obviously, for the derivation using the "integral" derivation, the following definition has been used : $R_{x} = \frac{1}{2\pi}\int_{0}^{2\pi}cos(\omega{t})cos(\omega(t-\tau))dt$ as the integral of a (co)sine is not defined over infinity. However, I do not get how it should affect the "fourier" derivation ? Dec 14 asked Autocorrelation derivation using fourier transform Aug 3 answered odd person out game Jul 31 revised Solving $5^n > 4,000,000$ without a calculator edited body Jul 31 comment Solving $5^n > 4,000,000$ without a calculator Oww ... I wanted to state that $10^7 > 4.10^6$ ... Thanks for pointing this out ! :> Jul 31 comment Solving $5^n > 4,000,000$ without a calculator may you explain the comment please ? Something bad in the answer ? Jul 31 awarded Editor Jul 31 revised Solving $5^n > 4,000,000$ without a calculator added 241 characters in body Jul 31 revised Solving $5^n > 4,000,000$ without a calculator latex correction Jul 31 answered Solving $5^n > 4,000,000$ without a calculator