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March 28th, 2016, 05:35 PM  #1 
Member Joined: Mar 2015 From: Los Angeles Posts: 73 Thanks: 7  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. $$ \left[ \lim{\frac{\sin x}{x}} \right]^2 = 1^2 = 1 = \lim {\frac{\sin^2 x} {x^2} } = \lim { \frac{1  \cos^2 x} {x^2} } = \lim \left[ (1 + \cos x) \cdot \frac{ 1  \cos x } { x^2 } \right] = 2 \cdot \lim {\frac{1  \cos x}{x^2}} $$ basically. 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 $\frac{1}{2} x^2$. 
March 29th, 2016, 08:22 PM  #2  
Senior Member Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications 
Hi mike1127, I like this, thanks. Quote:
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: $\displaystyle \large \sin^2 x+(1\cos x)^2=x_{approx}^2$ $\displaystyle \large \sin^2 x +1 2\cos x + \cos^2 x = x_{approx}^2$ $\displaystyle \large 22\cos x = x_{approx}^2$ $\displaystyle \large 2(1\cos x)=x_{approx}^2$ $\displaystyle \large \frac{1\cos x}{x_{approx}^2}=\frac{1}{2} \qquad$ 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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