For example, limit of cos(x)/sin(x) as x approaches pi from right/left... how does one do it, rigourously, instead of just evaluating the function at x = something just to the left of Pi, and x = something just to the right of pi?
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You can effectively consider how behave numerator and denominator when $x$ approaches $\pi$ and conclude.
Another way is to consider Taylor series built at $x=\pi$. You have $$\cot(x)=\frac{1}{x-\pi }-\frac{x-\pi }{3}+O\left((x-\pi )^3\right)$$ from which you can conclude.
I am sure that you can take from here.
If you consider Taylor series built at $x=\pi$ for numerator and denominator , you have $$\cos(x)=-1+\frac{1}{2} (x-\pi )^2+O\left((x-\pi )^4\right)$$ $$\sin(x)=-(x-\pi )+\frac{1}{6} (x-\pi )^3+O\left((x-\pi )^4\right)$$ to arrive to the same conclusions.
Claude Leibovici
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Strictly spoken, this is a Laurent series; a Taylor series does not allow negative powers of $x$. – AlexR Sep 18 '14 at 10:24
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@AlexR. You are very correct ... but it could just be a question of semantics. Let me say that I computed the Taylor series of $(x-\pi) \cot(x)$ at $x=\pi$. – Claude Leibovici Sep 18 '14 at 10:27
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That's of course also valid, but I suppose you knew we had a pole of first order at $x=\pi$ beforehand so that $(x-\pi)\cot x$ is defined at $x=\pi$. – AlexR Sep 18 '14 at 10:28
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@AlexR. Again, you are totally right. It is always good to argue. This is how we arrive to *Fiat lux*. Cheers – Claude Leibovici Sep 18 '14 at 10:39
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So our limit is:
$$\lim_{x\to\pi} \frac{\cos x}{\sin x} = \lim_{x\to\pi} \cot x$$
From the right:
$$\lim_{x\to\pi^{+}} \cot x$$
As $x\to\pi^{+}$, $\cot x \to\infty$
And as $x\to\pi^{-}$, $\cot x \to-\infty$
Therefore, since the left and right limits don't exist, the limit doesn't exist.
Prove this using the graph of $\cot x$.
AlexR
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Varun Iyer
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