
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 642 Thanks: 99 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 ).  

Tags 
geometrical, intuition, limits 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Intuition behind differential element dA?  triplekite  Differential Equations  3  June 11th, 2013 04:22 PM 
Vector Intuition check  taylor_1989_2012  Algebra  1  March 14th, 2013 08:26 AM 
A better intuition?  Sefrez  Calculus  1  August 16th, 2011 03:17 PM 
Very simple, just need some intuition.  Sefrez  Calculus  1  February 24th, 2011 02:58 AM 
Intuition Question (multivariable)  Billiam411  Calculus  4  October 27th, 2007 07:39 PM 