Difference between revisions of "Introduction to limits"
(Created page with "In many calculus textbooks, introduction to limits often starts with examples of limits such as <math>\lim_{x\to 2} x^2 </math> with invitation to evaluate the function, <math...") |
m |
||
| (10 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| + | |||
In many calculus textbooks, introduction to limits often starts with examples of limits such as <math>\lim_{x\to 2} x^2 </math> with invitation to evaluate the function, <math> x^2 </math> in this example, at points near <math> a </math> (so <math> x = </math> 2.1, 2.01, 1.9, 1.99 and so on). However, this type of examples is not particularly engaging and in fact may raise some questions as of why one needs to even consider limits in the first place. It would be more beneficial to introduces limits using examples in which the answer is at least not obvious. Another classical example is <math> \lim_{x\to 0} \frac{\sin x}{x} </math>. There are two problems with this one: it appears out of nowhere (i.e. as it presented is not motivated) and it is non-trivial to compute. The way around it is outlined below. | In many calculus textbooks, introduction to limits often starts with examples of limits such as <math>\lim_{x\to 2} x^2 </math> with invitation to evaluate the function, <math> x^2 </math> in this example, at points near <math> a </math> (so <math> x = </math> 2.1, 2.01, 1.9, 1.99 and so on). However, this type of examples is not particularly engaging and in fact may raise some questions as of why one needs to even consider limits in the first place. It would be more beneficial to introduces limits using examples in which the answer is at least not obvious. Another classical example is <math> \lim_{x\to 0} \frac{\sin x}{x} </math>. There are two problems with this one: it appears out of nowhere (i.e. as it presented is not motivated) and it is non-trivial to compute. The way around it is outlined below. | ||
Suppose you want to determine wether your planet is spherical or flat. You can draw a circle of some radius <math> r </math> and compare the circumference of this circle to its radius and analyze what happens with the circumference divided by the radius. | Suppose you want to determine wether your planet is spherical or flat. You can draw a circle of some radius <math> r </math> and compare the circumference of this circle to its radius and analyze what happens with the circumference divided by the radius. | ||
'''Plane''' | '''Plane''' | ||
| + | |||
| + | [[File:Circle-plane.png|alt=Circle on the plane|thumb|141x141px|Circle on the plane]] | ||
Let <math>C_r</math> denote the circumference of the circle of radius <math>r</math>. Then <math>C_r= 2\pi r</math> and therefore <math>\frac{C_r}{r}= 2\pi</math>; in particular this ratio does not depend on <math>r</math> | Let <math>C_r</math> denote the circumference of the circle of radius <math>r</math>. Then <math>C_r= 2\pi r</math> and therefore <math>\frac{C_r}{r}= 2\pi</math>; in particular this ratio does not depend on <math>r</math> | ||
| Line 8: | Line 11: | ||
'''Sphere''' | '''Sphere''' | ||
| − | Note that we measure the radius of the circle along the surface of the planet as the shortest distance from the centre to the circle. Let | + | [[File:Circle on the sphere.png|alt=Circle on the sphere|thumb|Circle on the sphere|226x226px]] |
| + | |||
| + | Note that we measure the radius of the circle along the surface of the planet as the shortest distance from the centre to the circle. Let <math>R </math> be the radius of the sphere. Note that <math>C_r = 2\pi {\color{red} a}</math> where <math>a=r\sin \theta</math> (see the diagram). Also, it is easy to see that <math>r = R\theta</math>. | ||
Therefore | Therefore | ||
| − | <math>\frac{C_r}{r} = \frac{2\pi a}{r} = \frac{2 \pi R\sin\theta}{R\theta} = 2\pi \frac{\sin \theta }{\theta}</math> | + | |
| + | <math>\frac{C_r}{r} = \frac{2\pi a}{r} = \frac{2 \pi R\sin\theta}{R\theta} = 2\pi \frac{\sin \theta }{\theta}.</math> | ||
| + | |||
| + | We see that the ratio of circumference to the radius depend on angle <math> \theta </math>. When <math> \theta </math> is small, the piece of sphere enclosed by our circle is "almost flat", so we should expect that <math> C_r /r </math> is very close to <math> 2\pi </math>. However, <math> C_r/r = 2\pi (\sin \theta /\theta) </math>. | ||
| + | Therefore we should expect that when <math> \theta </math> is close to <math> 0 </math>, <math> \frac{\sin \theta}{\theta} </math> is close to <math> 1 </math>. This suggests that | ||
| + | |||
| + | <math> \lim _{\theta \to 0}\frac{\sin \theta}{\theta} = 1. </math> | ||
| + | |||
| + | This observation can also be restated as <math> \sin \theta \approx \theta </math> when <math> \theta </math> is near <math> 0 </math>. As application, let us compute an approximation for <math> \sin 1 ^\circ </math>. Note first that we need to convert <math> 1 ^\circ </math> to radians: | ||
| + | |||
| + | <math> 1^\circ = \frac{\pi}{180} \rm{rad}.</math> | ||
| + | |||
| + | Therefore <math> \sin 1^\circ = \sin (\pi /180) \approx \pi /180 = 0.0174444\dots </math> (calculator gives <math> \sin 1^\circ = 0.017452406\dots </math>) | ||
Latest revision as of 21:49, 23 November 2021
In many calculus textbooks, introduction to limits often starts with examples of limits such as [math]\displaystyle{ \lim_{x\to 2} x^2 }[/math] with invitation to evaluate the function, [math]\displaystyle{ x^2 }[/math] in this example, at points near [math]\displaystyle{ a }[/math] (so [math]\displaystyle{ x = }[/math] 2.1, 2.01, 1.9, 1.99 and so on). However, this type of examples is not particularly engaging and in fact may raise some questions as of why one needs to even consider limits in the first place. It would be more beneficial to introduces limits using examples in which the answer is at least not obvious. Another classical example is [math]\displaystyle{ \lim_{x\to 0} \frac{\sin x}{x} }[/math]. There are two problems with this one: it appears out of nowhere (i.e. as it presented is not motivated) and it is non-trivial to compute. The way around it is outlined below. Suppose you want to determine wether your planet is spherical or flat. You can draw a circle of some radius [math]\displaystyle{ r }[/math] and compare the circumference of this circle to its radius and analyze what happens with the circumference divided by the radius.
Plane
Let [math]\displaystyle{ C_r }[/math] denote the circumference of the circle of radius [math]\displaystyle{ r }[/math]. Then [math]\displaystyle{ C_r= 2\pi r }[/math] and therefore [math]\displaystyle{ \frac{C_r}{r}= 2\pi }[/math]; in particular this ratio does not depend on [math]\displaystyle{ r }[/math]
Sphere
Note that we measure the radius of the circle along the surface of the planet as the shortest distance from the centre to the circle. Let [math]\displaystyle{ R }[/math] be the radius of the sphere. Note that [math]\displaystyle{ C_r = 2\pi {\color{red} a} }[/math] where [math]\displaystyle{ a=r\sin \theta }[/math] (see the diagram). Also, it is easy to see that [math]\displaystyle{ r = R\theta }[/math]. Therefore
[math]\displaystyle{ \frac{C_r}{r} = \frac{2\pi a}{r} = \frac{2 \pi R\sin\theta}{R\theta} = 2\pi \frac{\sin \theta }{\theta}. }[/math]
We see that the ratio of circumference to the radius depend on angle [math]\displaystyle{ \theta }[/math]. When [math]\displaystyle{ \theta }[/math] is small, the piece of sphere enclosed by our circle is "almost flat", so we should expect that [math]\displaystyle{ C_r /r }[/math] is very close to [math]\displaystyle{ 2\pi }[/math]. However, [math]\displaystyle{ C_r/r = 2\pi (\sin \theta /\theta) }[/math]. Therefore we should expect that when [math]\displaystyle{ \theta }[/math] is close to [math]\displaystyle{ 0 }[/math], [math]\displaystyle{ \frac{\sin \theta}{\theta} }[/math] is close to [math]\displaystyle{ 1 }[/math]. This suggests that
[math]\displaystyle{ \lim _{\theta \to 0}\frac{\sin \theta}{\theta} = 1. }[/math]
This observation can also be restated as [math]\displaystyle{ \sin \theta \approx \theta }[/math] when [math]\displaystyle{ \theta }[/math] is near [math]\displaystyle{ 0 }[/math]. As application, let us compute an approximation for [math]\displaystyle{ \sin 1 ^\circ }[/math]. Note first that we need to convert [math]\displaystyle{ 1 ^\circ }[/math] to radians:
[math]\displaystyle{ 1^\circ = \frac{\pi}{180} \rm{rad}. }[/math]
Therefore [math]\displaystyle{ \sin 1^\circ = \sin (\pi /180) \approx \pi /180 = 0.0174444\dots }[/math] (calculator gives [math]\displaystyle{ \sin 1^\circ = 0.017452406\dots }[/math])