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About 15 minutes ago I came across a question on MSE asking about $$\lim_{x\rightarrow 0} \left(\frac{9}{x} - 9\cot(x)\right)$$ Four people instantly answered it - two of the solutions used L'H$\hat{\mathrm{o}}$pital's rule, and the other two used Taylor series. I was wondering if there was a method of proving that the limit is zero using only trigonemtric identities and no powerful tools - then I came across this incredibly brief proof.

Let $x \in \mathbb{R}$. Then $$\frac{9}{-x} - 9\cot(-x) = -\frac{9}{x} + 9\cot(x) = -\left(\frac{9}{x} - 9\cot(x)\right)$$ As the function is odd, $\lim_{x\rightarrow 0}\left(\frac{9}{x} - 9\cot(x)\right) = 0$.

I'm certain I've overlooked something, because this is too good to be true. Where did I go wrong?

Harambe
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    https://math.stackexchange.com/questions/2513327/limit-as-x-approaching-0-of-9-x-9cotx/2513358#2513358 was without L'Hospital & Taylor's expansion – lab bhattacharjee Nov 10 '17 at 06:35

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Alright this is embarrassing but I just figured it out as I made the post - I need to know the limit exists before I can make such an assertion: $$f(x) = \sin(1/x)$$ is an odd function, but it clearly doesn't approach any limit as $x$ tends to $0$. If I can show that the limit exists, then it must be zero. There are plenty of odd functions which don't have limits at zero.

Harambe
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