Question

Concerning 1.Differentiation, Part A, Session 8
Clip 2: Limit of [1-cos(x)]/x

I learned in this lesson that lim_θ->0 (1 - cosθ)/θ = 0. Since both the numerator and the denominator is approaching 0, I gathered that the the numerator has to be approaching 0 faster than the denominator (in other words a higher rate of change). Can someone explain why does the numerator has a higher rate of change than the denominator?

Answer 1

I am pretty sure there is no answer to the question of "why" here.

I generally prove this by relying the previous result lim x ->0 sin(theta)/theta = 1. That in turn is proved by showing that sin(theta)/theta is trapped between two functions which are both approaching 1. This is called "squeezing" but was called "pinching" when I was a lad.

If you have that result, you can multiply the denominator and numerator of (1-cos(theta)/theta by 1+cos(theta). This makes it -sin^2(theta)/(theta(cos(theta) + 1) so you can express this as (-1)(sin(theta)/theta)(sin(theta)(some junk). However limit of the product is the product of the limits so it is lim(-1) x lim(sin(theta)/theta) x lim(sin(theta) x lim(some junk).

Since you know sin(theta)/theta has a limit of 1 you can replace that limit with 1. Then there is no factor of theta in any of the other denominators. Moreover, you can solve lim sin(theta) by substitution, and lim sin(theta) is 0. This makes one of the factors zero and no theta in the denominators so the limit of everything including the junk must be zero.

The reason for doing sin(theta)/theta first is that there is a relatively clear geometric argument that makes it seem plausible. I don't know of similar geometry argument for 1-cos(theta)/theta, so you do have to rely on the algebra. It seem like a trick, but the differential calculus will get even trickier, because if you look back at the difference quotient definition of derivative, derivatives ALWAYS involve limits in which both the numerator and denominator seem to be going to 0/0. So you have use anything you have got to rid the dominator of the zero tending term. Sometimes there is a good diagram for what you are doing; other times it is just blind hairy algebra.

Clips 1 & 2 of session 8 show both these limits worked.

Answer 2

The attached svg shows some 1st quadrant relations. Versine(x) is another name for 1 - cos(x). Perhaps you can find a pinching or other argument that is convincing.

Answer 3

Thanks :)

Answer 4

It also helps if you can imagine it this way.

Hopefully you have understood the intuition behind why \[\lim_{x \to 0} \frac{ \sin \theta }{ \theta } = 1\]
I'll be basing my argument on this.

When you recall the professor's diagram. As the line sine line approaches closer and closer to the arc length. The quantity which is described as a versine in Lar's SVG also becomes smaller and smaller.

This is how I convinced myself that this limit is true.

Answer 5

because if the denominator is approaching 0 faster than the numerator, then, eventually the denominator will become smaller than the numerator, then numerator/denominator > 1 and will approach infinity as x approaches 0.
for ex: -
f(x) = x/(x^2)
(x is approaching 0 )

numerator denominator Quotient
x x^2 f(x)
---------------------------------------------
10 100 0.1 (numerator < denominator)
1 1 1 (numerator = denominator)
0.1 0.01 10 (numerator > denominator)
(denominator approaches 0 faster than the numerator hence the quotient approaches infinity)

If the numerator and the denominator approach 0 at a constant rate then the quotient will approach a constant.

Hence numerator has to approach 0 faster than the denominator for the quotient to approach 0. Hope this helps.