Why you gotta be so complicated

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?

Easy Math Editor

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`2^{34}`	$2^{34}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

With $\cos(x) = 1 - \dfrac{x^{2}}{2} + O(x^{4})$ 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 - 5 years, 4 months ago

Haha, thanks for making this note less lonely :3.

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 - 5 years, 4 months ago

I tend to use L'Hopital's as a last resort. It is very useful but it feels like cheating when there is an alternative approach available, and on those occasions that it does not work efficiently, those alternative methods might come in handy. :)

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 - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$