Question

I don't understand why sin delta x over delta x equals 1. This is from the 3rd lecture. Anyone get it?

Answer 1

Is there a limit involved??

Answer 2

THe lim as x approaches infinity of sin(x)/x =1. That is one of the most fundamental limits in all of calculus. The reason that this limit is so important is because when you use a rule called l'hopital's rules (Which is taking the derivative of the numerator and denominator) which comes out to be lim as x approaches infinity of f^1(X)=cos)x)/1 which is 0!!!!! However the limit actually is 1 because of the limit.

Answer 3

He proves it geometrically.

If you draw a right triangle of the unit circle in the 1st quadrant and label the angle at the center 'x'. Sin x = length the opposite side (because sin = opp / hyp and hyp = 1 in unit circle) and x = arclenth (length of the circle that it subtends).

Now, as you shrink angle x, sin x approaches arclength x. Which means that as angle x approaches 0, sin x = x (arclength), therefore sinx/x = 1.

Draw it out and hopefully you can see it for yourself.

Answer 4

well "r" is the radius in above figure.
i will prove it by considering the fact
area of triangle OBD < area of cake piece OBD < area of triangle OBC
1/2 OB x AD < 1/2 r^2 x < 1/2 OB x BC
note ;
AD=OD sinx ,BC=OB tanx
plug these values
OB x OD sinx < r^2 x < OB x OB tanx
note;
OB = OD = radius r
r^2 sinx < r^2 x < r^2 tanx
playing with these inequalities we get
1 < x/sinx < 1/cosx
i can rewrite it as
1 > sinx /x > 1/cosx

apply limit to 1/cosx as x goes to zero so we get 1
so 1 > sinx / x >1
by handwitch theorem sinx /x =1

Answer 5

x is in radians...

Answer 6

Usman is correct using the squeeze theorem. What I was suggesting is that (when using Usman93's drawing), as angle x shrinks to zero, side AD (Sin x) approaches arclength BD (x). As one approaches the other the ratio (sin x /x ) approaches 1.

Answer 7

You might also look up the (closely related) "small angle approximation" or "sinc function" for pictures/other perspectives.

Answer 8

Well,skip to see the literature 35 about the L`Hôspital`s Rule and you will find out that the sin(x)/x as x goes to 0 is 1 because sin(0)=0,so we take both of the derivatives of them and that is cos(x)/1.Plug in x=0 then it is equal to 1.The rule will explain that well.There are always some strange things in the first literatures since Jerison could not explain then.Just keep on studying and you will get them in the future

Answer 9

@Xavi: You cannot use L'Hôspital's rule to define the limit sin(x)/x when x approaches 0 before you define what is the derivative of sin(x) and to define the derivative of sin(x) you need to know what is the limit of sin(x)/x as x approaches 0. This is circular definition and as such invalid.

You can use squeeze theorem on sin(x)/x (x in radians) as follows:
\[\sin x < x <\tan x \rightarrow
\sin x < x < \frac{ \sin x }{ \cos x }\rightarrow
1<\frac{ x }{ \sin x }<\frac{ 1 }{ \cos x }\rightarrow1>\frac{ \sin x }{ x }>\cos x\]Now because cos(0) is 1 it follows that the limit of sin(x)/x as x tends to 0 is also 1.