Infinity is a sketchy topic, for example \(\infty \neq \infty\).

In the expression, \(\dfrac{\left(\frac{1}{x}\right)}{\left(\frac{1}{1-\cos(x)}\right)}\).

As \(x\) tends 0, we get \(\displaystyle \lim_{x\Rightarrow 0} \dfrac{\left(\frac{1}{x}\right)}{\left(\frac{1}{1-\cos(x)}\right)}=\dfrac{\infty}{\infty}=0\).

However, in the expression, \(\displaystyle \lim_{x\Rightarrow 0} \dfrac{1-\cos(x)}{x}=\dfrac{0}{0}=0\)

In the first equation, \(\dfrac{\infty}{\infty}=0\), is the infinity in the numerator any smaller than the infinity in the denominator?

Similarly, in the second expression, is the 0 in the numerator smaller than the 0 in the denominator? Or is neither equivalent to 0?

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TopNewestWith \(\cos(x) = 1 - \dfrac{x^{2}}{2} + O(x^{4})\) we have \(\dfrac{1}{1 - \cos(x)} = \dfrac{2}{x^{2} + O(x^{4})}\), which goes to \(\infty\) "faster" than \(\dfrac{1}{x}\) does as \(x \rightarrow 0\). Similarly \(1 - \cos(x) = \dfrac{x^{2}}{2} + O(x^{4})\) goes to \(0\) faster than \(x\) does as \(x \rightarrow 0\). Of course the cleanest way to solve the limit in the second form is to note that

\(\lim_{x \rightarrow 0} \dfrac{1 - \cos(x)}{x} = \lim_{x \rightarrow 0} \left(\dfrac{1 - \cos(x)}{x} * \dfrac{1 + \cos(x)}{1 + \cos(x)}\right) = \lim_{x \rightarrow 0} \dfrac{\sin^{2}(x)}{x(1 + \cos(x))} =\)

\(\lim_{x \rightarrow 0} \sin(x) * \lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} * \lim_{x \rightarrow 0} \dfrac{1}{1 + \cos(x)} = 0 * 1 * \dfrac{1}{2} = 0\).

Note that \(\lim_{x \rightarrow 0} \dfrac{x}{1 - \cos(x)}\) does not exist since the limit from the left is \(-\infty\) and the limit from the right is \(\infty\).

I figured that you knew all of this already, but your note seemed lonely so I thought it deserved the company of a response. :) – Brian Charlesworth · 8 months ago

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Yes, I know how to evaluate these limits as they approach 0, although I've never used these creative approaches. I've always just used l'hopital's or much simpler rearrangements (that don't always work).

I think I should reword what I was asking. Can it be said that \(\displaystyle \lim_{x\rightarrow 0} \frac{1}{x}< \frac{1}{1-\cos(x)}\) (is \(\infty<\infty\))? When you cross multiply we get \(1-\cos(x)=x\), which has one solution of \(x=0\). However, when we leave it in fractional form we have \(\frac{1-\cos(x)}{x}<1\) which is also true. – Trevor Arashiro · 8 months ago

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Your first inequality is true, but note that when we flip it we would have \(x \gt 1 - \cos(x)\), which is true only for \(x \gt 0\). It is the reverse for \(x \lt 0\), (and as you note \(x = 1 - \cos(x)\) for \(x = 0\)). This intuitively suggests that the limit of one of \(\frac{x}{1 - \cos(x)}\) or \(\frac{1 - \cos(x)}{x}\) is going to be \(0\) and the other is not going to exist, and because of the validity of your first inequality we can be comfortable in guessing that \(\dfrac{\frac{1}{x}}{\frac{1}{1 - \cos(x)}} = \frac{1 - \cos(x)}{x}\) is going to be the one that goes to \(0\). – Brian Charlesworth · 8 months ago

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