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March 28th, 2016, 05:35 PM   #1
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geometrical intuition on some limits

I wanted to see if I could produce a geometrical diagram (also wanted to just fool around with Asymptote) that gave some intuitive insight into two particular limits:

$$ \lim_{x \to 0}{\frac{\sin x}{x}} = 1 $$

$$ \lim_{x \to 0}{\frac{1 - \cos x}{x^2}} = \frac{1}{2} $$

What's fascinating to me is that $\sin x$ seems to "go to zero slower" than $1 - \cos x$, as the former is an equivalent infinitessimal to $x$, while the latter to $x^2$. The latter can be proved from the former as follows.

$$ \left[ \lim{\frac{\sin x}{x}} \right]^2 = 1^2 = 1 = \lim {\frac{\sin^2 x} {x^2} } = \lim { \frac{1 - \cos^2 x} {x^2} } = \lim \left[ (1 + \cos x) \cdot \frac{ 1 - \cos x } { x^2 } \right] = 2 \cdot \lim {\frac{1 - \cos x}{x^2}} $$


Now for a picture. This is an arc of a unit circle of size x radians. The chord labeled $ \approx x $ is close to the arc length which is $x$ (because this is a unit circle and $x$ is in radians).

So we can see that as $x \to 0$, the little triangle will get narrower and narrower, so $\sin x$ and $x$ will approach each other. (This is just an intuitive argument.) But notice how small $1 - \cos x$ is by comparison. It will become a tiny strip as the triangle becomes very narrow. So to find an equivalent infinitessimal we need to choose something that goes to zero a bit faster than $x$, and it turns out to be $\frac{1}{2} x^2$.
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March 29th, 2016, 08:22 PM   #2
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Hi mike1127,

I like this, thanks.

(This is just an intuitive argument.)
Good, I'm not very good at proofs anyway, so with this out in the open, maybe the mathematicians that would normally demand rigor (rightly so) might let me continue to play fast and loose with the rules.

In your diagram, I would like to look at the right triangle formed by $\sin(x), \ 1-\cos(x), \ \text{ and } \approx x$ which I will call $x_{approx}$ . Then it seems that one may write:

$\displaystyle \large \sin^2 x+(1-\cos x)^2=x_{approx}^2$

$\displaystyle \large \sin^2 x +1 -2\cos x + \cos^2 x = x_{approx}^2$

$\displaystyle \large 2-2\cos x = x_{approx}^2$

$\displaystyle \large 2(1-\cos x)=x_{approx}^2$

$\displaystyle \large \frac{1-\cos x}{x_{approx}^2}=\frac{1}{2} \qquad$ and since $x \approx x_{approx}$

$\displaystyle \large \frac{1-\cos x}{x^2}=\frac{1}{2}$

Also, if we look at the angle I'll call $\theta$ between the vertical $\sin x$ line and the $\approx x$ line:

$\displaystyle \large \sin \theta = \frac{1-\cos x}{x_{approx}}=\frac{1-\cos x}{x_{approx}^2} \cdot x_{approx}=\frac{1}{2} \cdot x_{approx}=\frac{x_{approx}}{2} \approx \frac{x}{2}$

Since $\theta$ and $x$ are small:

$\displaystyle \large \theta \approx \frac{x}{2}$

So it seems to me that your diagram also intuitively explains the $1-\cos x$ limit too: The large triangle with an angle of $x$ and a hypotenuse of 1 creates another triangle with an angle of $x/2$ and a hypotenuse of $\approx x$ giving $\approx x^2/2$ .

I hope that you agree, and I never would have thought about this without your post (and mathematicians please look away ).
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